 So, hello students. Good afternoon. So, today we will take the exercise six of sequence and very chapter. So, in this exercise, we will be particularly dealing with the questions on series. Like questions based on method of submission, whether it be a Sigma method of submission or method of difference, including VN method. And also, we will see the questions based on arithmetic or geometric progressions, like we what we used to call it as a GP. So, without wasting any time, let's start our working on exercise number six. So, here's the first question. We're saying the sum of first in terms of the series, one series is given here 1 by 2 plus 3 by 4 plus 7 by 8 plus 15 by 16. And we have to find the sum of first in terms of this series. So, let's try to write the Rth term of this series. Okay, we will write the Rth term of this series. So, what will be the Rth term, if you observe it will in denominator, it's a simple right in denominator, it's a two raised to power R. Two raised to power R. Okay. And in numerator, it is two raised to power R minus one, if you can observe, it is two raised to power R minus one upon two raised to power R. So, this is our Rth term for this particular series. Anyway, anytime we, if we have any doubt we can put the value of R and we can check whether our Rth term what we have written is correct or not. So, for time being let's I'm putting it as R is equal to three. So, our third term will be two raised to power three minus one upon two raised to power three. So, it will be eight minus one that is seven by eight. So, here also our third term is seven by eight, hence our Rth term what we have written is correct. Okay. Now, we have to find the sum of first in terms of this series. So, it will be sigma TR R going from one to N. Okay. So, we write it as sigma R is equal to one to N. What is our Rth term? It is two raised to two raised to power R minus one upon two R. We can further rewrite our Rth term as one minus one upon two raised to power R. Right. So, this will be summation R going from one to N. So, now I will distribute this sigma over both the terms and it will become one R raised to power one. Sorry, R going from one to N minus one upon two raised to power R, R is going from one to N. So, this thing will become N one plus two plus three. Okay. This will become N because one, it is not having any R component in it. So, it will be one plus one plus one, how many times? N times. So, it will be basically N and what will be this? This will two raised to power one put R equal to one. It will be one by two plus one by two square plus one by two cube up to one upon two raised to power N. So, it is basically a GP. This is a GP where the first term is, where the first term is one by two and common ratio R is also one by two. So, we can write the sum of N terms of a GP. It will be a R raised to power N or we can say one minus R raised to power N upon one minus R, one minus R. Right? So, further we can rewrite it as N minus this one by two and this one by two, this will become one by two in the denominator. So, it will be cancelled out. So, it will become N minus one plus one upon two raised to power N. Further, we can write this as plus two raised to power minus one. So, two raised to power minus N plus N minus one will be our answer. Two raised to power minus N plus N minus plus N minus one. So, option C. Option C is correct here. So, this question is done. Let's move to the question number two. So, here one product is given two raised to power one by four into four raised to power one by eight into eight raised to power one by 16 into 16 raised to power one by 32. So, this product we can write it as two raised to power one upon two square, right? One by four can be written as one upon two raised to power two. Then we can write it as two square four can be written as four, we can write four as two square and in power it is one by eight. So, we can write it as one upon two cube. Similarly, this eight we can write as two raised to power three or two cube. Power may two raised to power four. Then two raised to power four. Escape power may a jagat two raised to power five. And this product will go up to infinity, right? Sorry. So, if you see, okay, we will go one step further and we will write it as two, nothing is there. So, it is power one only and power will be multiplied by this. So, it will be one by two square as it was earlier. But look at this term. It will be two raised to power two upon two raised to power three, right? And it will become two raised to power three upon two raised to power four. Similarly, this term will become two raised to power four upon two raised to power five. Okay, this will continue up to infinity. Now, the target behind this is now you can see in this the base term is same like the base term is two only. So, we can write it as two and this power will be summed up. Like we can sum the power of this. So, it will become two raised to power two one upon two square plus two upon two cube plus three upon two raised to power four plus four upon two raised to power five. And this will go up to infinity. So, now what we will do, let's assume this as a series. What's written in the power of two. We will take it as a series s. So, we can write s as one upon two one upon two square plus two upon two cube plus three upon two raised to power four plus four upon two raised to power five dot dot dot dot up to infinity. Okay. Now, if you observe, sorry, if you observe this is the terms written in numerator is in EP 1234 with a common difference of one and that is multiplied by a GP with the common ratio one by two. So, for this what we do we use to we use to multiply this sum by the common ratio that is one by two s and we use to write the terms by shifting one side, like by shifting by right hand side. So, we can write it as it will become one upon two cube. Then it will become two upon. So, two upon two raised to power four, then three upon two raised to power five, and it will go up to infinity. Now, we will subtract this. We will subtract. So, subtracting this, this minus this we will have s by two, right. s minus one by two s will be s by two is equal to one by two squared plus one by two cube plus one by two raised to power four plus one upon two raised to power five and up to this will continue up to infinity. So, again, this is this. This thing is forming a GP, right. This is a GP with a common ratio of one by two and the first term is one by two squared or one by four we can say. So, we can write it as and this is going up to infinity. So, this is a infinite geometric progression series. So, it's a sum can be written as a upon one minus sum of infinite series of a GP, right. So, what is a? A is one upon four and one minus r, r is half. So, one by four upon one by two. So, it will become one by two. So, from here, our s becomes one, right. So, what's written in the power of two, this whole thing becomes one, this whole thing becomes one. Sorry, right. So, this s is becoming one. So, our product will be finally our product will be two raised to power s, what we have considered this two raised to power s. So, it will be two raised to power s, what we got s is equal to one. So, two raised to power one is equal to two. So, our product will be two. This will be our answer. So, option C is correct. Option C is correct for this question. Now, going ahead, let's move to question number three. One series is written here 1 plus 3 plus 7 plus 15 plus 31 plus dot dot dot up to n terms. So, we have to find the value of the series up to n terms. Okay, let's do it. So, let's assume this is some series as h, this summation as s. So, it will be 1 plus 3 plus 7 plus 15 plus 31 plus dot dot dot up to n terms. So, we have to go up to n terms, right. Now, if you observe the difference of the terms, the difference of the terms is will attain one progression. How? See, 3 minus 1 is 2, 7 minus 3 is 4, 15 minus 7 is 8, 31 minus 15 is 16, right. So, the difference of the term, the first difference of the terms, this we are getting a GP. We are getting a GP. What is the common ratio here? Common ratio is 2, okay. So, for such type of series, we can write the nth term as, sorry. So, we can write the nth term for such type of series as a r raise to power n minus 1 plus bn plus c, right. What is r here? The r is the common ratio, common ratio which is obtained after taking the difference. So, r here is 2, okay. So, our nth term, we can write the nth term of this original series as a 2 raise to power n minus 1 plus bn plus c. And what is a, b, c here? a, b, c look, a, b, c are constants here whose value we can identify, like whose value we can calculate, how to calculate, we will see it. So, we can write the nth term. Now, we have written, since we have written the nth term, now let's find the value of a, b and c. So, put n is equal to 1 here. So, we will get t1, t1 will be equal to a into 2 raise to power 0, 2 raise to power 0, that will become a anyhow, plus b plus c. So, I am deleting it and writing a. So, it will become a plus b plus c when we have taken n as 1. So, a plus b plus c is equal to, now what is the first term? First term is 1. So, we will equate it to 1. Now, let's write the second term, putting n equal to 2. So, it will become 2 raise to power 1 means 2a, 2a plus 2b plus c is equal to, what is the second term? Second term is 3. And similarly, let's write the third term as 2 raise to power 2, that will be 4a plus n was 3. So, 2 raise to power 2, 4a plus 3b plus c is equal to 7. Now, let's assume it as equation 1, this equation 2, this equation 3. So, subtracting equation 1 from equation 2, we will get a plus b is equal to 2. Similarly, subtracting 2 from 3, we will have 2a plus b is equal to 4. Now, again subtract it, again subtract it, we will have a is equal to 2. If a is equal to 2, put it its value here. So, if a is equal to 2, b will become 0. So, a equal to 2, b equal to 0, then put the value of a and b in equation 1, you will get 2 plus 0, 2 plus 0 plus c is equal to 1. So, c will become minus 2 plus 1, that will be minus 1. So, we got the value of a, b and c. So, we got the value of nth term also, right? So, we can, now we will write the nth term of the series as a was a is 2. So, 2 into 2 raise to power n minus 1 plus the value of b is 0. So, it will eventually become 0 and what is c? c is minus 1. So, the nth term of the series becomes 2 raise to power n minus 1 plus 1, that will be n and minus 1. So, this is our nth term. Now, we have to find the sum of n terms. So, simple you put sigma t r. If nth term is this, then our rth term will be replaced n by r, 2 raise to power r minus 1. So, sigma t r, where r is going from 1 to n. So, it will become sigma 2 raise to power r minus 1. r is going from 1 to n. So, distribute this sigma over both the terms. It will become 2 raise to power 1 plus 2 raise to power 2 up to 2 raise to power n and this will become minus n. So, this is a GP with first term as 2. So, we will write it as 2 r. What is r here? r is 2 here. r raise to power n means 2 raise to power n minus 1 upon r minus 1 that is 2 minus 1 minus n. So, it will become 2 raise to power n plus 1 minus 2 minus n. So, this is our summation of n terms. 2 raise to power n plus 1 minus 2 minus n. So, option b is correct. So, option b is correct for this question. Hope this is clear to all. Now, let's take the next question, question number 4. So, 99th term of the series, this series is given as 2 plus 7 plus 14 plus 23 plus 34 plus dot, dot, dot it and it is asking the 99th term, the value of 99th term. Okay. So, series given is 2 plus 7 plus 14 plus 23 plus 34 plus dot, dot, dot, dot. So, if you see, now let's take the difference. So, 7 minus 2 is 5, 14 minus 7 is 7, 23 minus 14 is 9, 34 minus 23 is 11. Okay. So, we are getting AP here after taking the first difference of the terms. So, this is actually AP, right? This is actually AP with the common difference of 2, with the common difference of 2. So, for such types of series, we write the nth term as, we write the nth term as a n square plus b n plus c, right? Where a, b, c are some constants, which we can anyhow calculate like we did in our last question. So, but the important thing is the nth term for such type of series can be given as a n square plus b n plus c means it will be a quadratic in n. So, let's try to find the value of a, b and c. For that, we will put n equal to 1 and we will find the value of t1. So, it will become a plus b plus c and what is t1 here? t1 is 2. Similarly, find t2. t2 will be 4a, 4a plus 2b plus c equal to 7. Second term is 7 here. The third term will be 9a, 9a plus 3b plus c is equal to 14. Third term is 14. Now, let's take as equation 1, equation 2 and equation 3. Subtract equation 1 from equation 2. We get 3a plus b is equal to 5. Same way, subtract the equation 2 from 3. It will become 5a plus b is equal to 5a plus b. c will be cancelled out. 14 minus 7 will be 7. Now, again subtracted, you will get 2a, b will be cancelled. 2a is equal to 2. From here we get a is equal to 1. Now, put value of a here. We will get if 3 means b will be 2. Now, put the value of a and b in equation 1. 1 plus 2, 3. 1 plus b is 2 plus c is equal to 2. So, c will become minus 1. c will become minus 1. So, we got the value of a, b and c means we got the nth term. What will be the nth term? nth term will be n square plus b and b and a was 1, no? And b was 2, okay. So, nth term will be n square plus 2n and c is minus 1. So, this will be the nth term of the given series, right? The nth term of this series. So, question is asking to find the 99th term, right? T99th. So, we have to put the value of ns99. So, let's do one thing. Let's make it a perfect square. So, it will be n square plus 2n plus 1 and minus 2, right? Now, the benefit of this will be it will become n plus 1 whole square minus 2. So, put, now put n as a 99 here. So, it will be 100 squared minus 2. That will be 10,000 minus 2. That will be nothing but 9,998. So, this will be our 99th term. So, option a. Option a is the correct one for this question. Okay. So, this question is done. Let's take our fifth question. So, this one is 6, this one is 5th. Okay. So, let's take our fifth term. The sum of the series 1 into 2 into 3 plus 2 into 3 into 4 plus 3 into 4 into 5 plus up to n terms is. Okay. So, first write the rth term for this series. First write the rth term of this series. So, rth term we can write it as r into r plus 1 into r plus 2, right? This will be the rth term put r is equal to 3 to cross verify. So, t3 will be 3 into 4 into 5 and here also t3 is 3 into 4 into 5 means rth term we have written correctly. Okay. So, we have to find the sum of this series. So, I will do one thing. I will do like we will solve this question by VN method. So, what I will do? I will try to break this rth term. I will try to break this rth term in terms of like in difference of two such functions which will be consecutive in r, right? I will break this rth term in such type of functions which will be consecutive in r. So, let's try to figure it out. So, I am defining one function VR as I am defining one function VR as where I will be taking one like r, r plus 1, r plus 2 into r plus 3. Like I have taken this extra term here, right? And I will also define VR minus 1, right? So, what will be VR minus 1? It will be r minus 1 into r then r plus 1 and r plus 2, right? Now, take the difference of these two functions. Taking the difference of these two functions, we will eventually see our the rth term. How? Let's do it first. Let's first do it. So, VR minus, VR minus 1, so VR minus VR minus 1 will become, so if you observe these three terms are common. These three terms are common in both the functions. So, I am taking that common that will be r into r plus 1 into r plus 2 and what is left out? Left out is r plus 3 minus r minus 1, right? So, it will become r into r plus 1 into r plus 2 and what will become r plus 3 minus r plus 1 means 4. Like it will become 4. Now, you observe what is written here. This r into r plus 1 into r plus 2 is nothing but our rth term, right? So, we can write it as 4 times of dr. This is what our target was. We had to break, like we were trying to break this rth term as a difference of two functions which are consecutive in r. So, we defined one function VR. We also defined one function putting r less than 1. So, VR minus VR minus 1 is equal to our 4 times rth term, right? So, this will be very much useful. How? We will see now. From here we can say our rth term will be 1 by 4 times VR minus VR minus 1, ok? Now, let's write our first term. So, our first term will be 1 by 4 put r is equal to 1 here. It will be v1 minus v0. What will be our second term? Second term will be 1 by 4 v2 minus v1. Similarly, third term will be 1 by 4 v3 minus v2 dot, dot, dot. If we take the nth term, nth term will become 1 by 4 vn minus vn minus 1, right? Now, submit, add all. Adding all, we will get t1 plus t2 plus t3 up to tn. This is nothing but our sum to the nth term is equal to 1 by 4. Now, observe this thing. This v1 is getting cancelled out with this minus v1. This v2 thing is going out with this minus v2. Similarly, this v3 will go with minus v3 and this vn minus 1 will also go from its preceding term. So, we are left with 1 by 4 into vn minus of v0. So, here we are left with vn and here we are having minus v0. So, our sum to n terms becomes 1 by 4 vn minus v0. Now, what is vn and v0? We have defined the function vr here. So, put n in place of r. So, we will get vn and put 0 in place of r in vr. We will get v0. So, it will become 1 by 4 and n into n plus 1 into n plus 2 into n plus 3 and minus v v0. v0 will be nothing but 0 only. So, write 0. So, finally, we are left with n into n plus 1 into n plus 2 into n plus 3 upon 4. So, this will be the sum to n terms. So, option c is correct. Option c is correct for this question. So, let's take the next question, question number 6. 1 upon 1 into 2 plus 1 upon 2 into 3 plus 1 upon 3 into 4 dot dot dot dot plus 1 upon n into n plus 1. So, here already the nth term is given as 1 upon n plus 1. So, we can write the rth term as 1 upon r into r plus 1. Here also we will apply the vr method or vn method. How? We will again break this tr in 2 such functions will be which will be consecutive in r. How? Let me define one function vr as 1 upon r plus 1. So, what will be our vr minus 1? vr minus 1 will be 1 by r. Replace r by r minus 1. It will become r and this will become r minus 1 means vr minus 1. No. So, replace it. No. r minus 1 let me replace. Okay. We have defined this as vr. So, replacing r by r minus 1 here we will get 1 by r only. Okay. So, now this vr minus 1 minus vr. Let's observe it. It will be 1 by r minus 1 upon r plus 1. Okay. So, it will become r into r plus 1. Then this will r plus 1 minus of r. So, it will be 1 upon r into r plus 1. Right. So, this is our vr minus 1 minus of vr. This is nothing but 1 by r into r into r plus 1 is tr. So, this is our tr or the rth term. Right. So, our rth term becomes vr minus 1 minus of vr. Now, let's identify. Let's find out the first term. So, our first term will be v0 minus v1. Our second term t2 will be v1 minus v2. Similarly, third term will be v2 v2 minus v3. Okay. Similarly, it will go up to nth term. So, nth term will be vn minus 1 minus of vn. Now add all this. Adding all this it will become sum 2 n terms. Now observe here. It will minus v1 plus v1 will be cancelled. Minus v2 plus v2 will be cancelled. Then minus v3 this thing will be cancelled out. So, we will be left out with v0 minus of vn. Okay. So, this will be our sum 2 n terms. Now sum 2, what is v0? v0 is put n equal to 0 here in this function. So, n equal to 0 will be v0 will be 1 and vn will be replace r by n. It will be n plus 1. So, our sum of n terms will become v0 is 1, vn is 1 upon n plus 1. So, n plus 1, n plus 1 minus 1 it will become n upon n plus 1. So, sum 2 n terms will be n upon n plus 1. This is our answer. So, option b is correct. Now let's take this seventh question. Sum of n terms of the series 3 upon 1 square plus 5 upon 1 square plus 2 square plus 7 upon 1 square plus 2 square plus 3 cube, 3 cube. Okay. There is one correction in this question. It will be 3 squared basically. So, let me correct. This will be 3 squared. Okay. So, we have to find the sum of n terms of this series. Okay. So, let me write, we can proceed with writing rth term and we will use sigma method of summation. Let's try it like our rth term will be, if you see our rth term will be the numbers in the numerator are basically ap with the common difference of 2 and the first term being 3. Okay. So, let me write it as like a plus n r minus 1 into d. Okay. a plus r minus 1 into d. Like I am finding the rth term of the ap written in the numerator. And in denominator if you see denominator it says 1 square plus 2 square plus 3 square dot dot dot up to r square. Okay. So, what is a here is 3 plus r minus 1 d is 2. And this thing 1 square plus 2 square plus 3 square up to r square we can write it as like sum of first n squares of first n natural numbers we write it as r into r plus 1 that 2 r 2 r plus 1 upon 6 right. So, this numerator will become 2 r plus 1. Okay. 2 r plus 1 and this will become r into r plus 1 2 r plus 1 upon 6 this will go to numerator and this 2 r plus 1 this 2 r plus 1 we cancelled out. So, t r means rth term in the simplest form we can write it as 6 upon r into r plus 1. Okay. Now, this question now seems to be easy because here once again we can apply that we in method what we did in last I think fifth or sixth question. So, this tier is basically 6 upon r into r plus 1. So, now let me define one function vr I am defining one function vr h 1 upon r plus 1. So, what will be vr minus 1 or vr minus 1 will be 1 upon r. Okay. Now, let's take the difference vr minus 1 minus vr it will become 1 upon r minus 1 upon r plus 1. So, this will eventually become 1 upon r into r plus 1 right and if you say 1 upon r into r plus 1, but our tier is 6. So, we can write it as 1 upon 6 into tier right. So, this vr minus 1 minus vr we can write this as 1 upon 6 into tier because if we put tier the value of tier 6 upon r into r plus 1. So, 6 and 6 will be cancelled out. So, it's okay. We can write our, so from here we can write our rth term as we can write our rth term as 6 times vr minus 1 minus vr. Okay. So, let's write the first term first term will be v0 minus v1 t2 will be 6 times v1 minus v2 t3 will be 6 into v2 minus v3. Similarly, nth term will be 6 times vn minus 1 minus vm. Okay. So, adding all these we will get sum to n terms as this v1 minus v1 plus v1 will be cancelled minus v2 plus v2 minus v3 plus v3. And so finally, we will get sns 6 times v0 minus vn. Okay. Now, what will be our v0? Our v0 will be put n equal to r equal to 0 here. We will get v0 as 1 upon 1 and our vn will be 1 upon n plus 1. So, our sum to n terms will become 6 times v0 is 1 minus vn is 1 upon n plus 1. So, 6 times n plus 1 this will be n plus 1 minus 1, 6n upon n plus 1. So, this will be the sum to n terms sn will be 6n upon n plus 1. Option C. So, option C is correct. So, we are done with this. Now, let's move to the next question. Question number 8. Question number 8. Right. So, one series is, okay, the value of nth term is given here as 1 by 4 into n plus 2 into n plus 3 for n equal to 1, 2, 3, dot, dot, dot. Then we have to find the value of 1 upon t1 plus 1 upon t2 plus 1 upon 3 t3 plus dot, dot, dot 1 upon. 2003 term. Okay. And nth term, the value of nth term, they have provided it. Okay. So, nth term is given as 1 upon 4 into n plus 2 into n plus 3. Okay. So, we can write 1 upon tns, we can write 1 upon tns, 4 upon n plus 2 into n plus 3, right. 4 upon n plus 2 into n plus 3 is equal to 1 upon nth term, 1 upon tn. Now, let's try to go through a VN method, because I am able to say that we can express this 1 by tns, the difference of two such functions which are consecutive in n or r. So, I am defining one function VN as 1 upon n plus 3. Okay. 1 upon n plus 3. So, what will be VN minus 1? VN minus 1 will be, replace n by n minus 1, it will be n plus 2. 1 upon n plus 2. Now, take the difference of these two. This will be VN minus 1 minus of VN is equal to 1 upon n plus 2 minus 1 upon n plus 3. Okay. So, it will become n plus 2 multiplied by n plus 3. In numerator, it will become n plus 3 minus n minus 2, right. So, this plus n minus n will be cancelled, 3 minus 2 will be 1. So, this will be 1 upon 1 upon n plus 2 into n plus 3, right. This is our VN minus 1 minus VN. And what was our nth term? nth term was 4 divided upon. This was our nth term. Like, I am taking 1 by tn as the like nth term since our question was asked in terms of 1 upon tn. So, we can write it as like 1 upon, let me divide it by 1 by 4. Let me divide it by 1 by 4. So, it will become 4 in the numerator. Okay. Right. No. So, eventually we can say this VN minus 1 minus VN is equal to 1 by 4 times. Now, we will rewrite it as 1 upon tn. Okay. Since 1 upon tn was given as like we have calculated it as 4 upon n plus 2 into n plus 3. So, this equation we will use further. So, our 1 upon tn will become, this becomes equal to 4 into VN minus 1 minus of VN, right. Our 1 upon tn, 1 upon t1 will be 4 into V0 minus V1. 1 upon t2 will be 4 into V1 minus V2. 1 upon t3 will be 4 into V2 minus V3. Similarly, 1 upon tn will be equal to 4 into VN minus 1 minus of VN, right. So, adding all this we will get the required sum what was asked in the question. So, it will be sum of N terms will be equal to 4 into these things will be cancelled out minus V3 plus V3. So, V0 minus of VN, right. And what will be V0? What will be V0? VN is this function. So, V0 will be put N equal to 0 there. So, it will be 1 upon 3 and our VN will be 1 upon n plus 3, right. So, our sum to N terms become 4 into V0 is 1 by 3 minus 1 upon n plus 3, okay. But let's see the question. Question is asking to find the value like 1 upon, it is saying to find the value of 2, 1 upon t2003 only, right. It is going up to, N is going up to 2003. So, let's put N equal to 2003. So, N equal to 2003 if we put our sum to 2003 terms, 2003 terms will become 4 into 1 by 3 minus 1 upon any 2003 plus 3. So, this will be 2006. So, finally it will become, let's take LCM 3 into 2006. So, 2006 minus 3, okay. So, it will be 4 into 2003 upon 4 into 2003 upon 3 into 2006, right. So, let's divide it by 2. It will become 2. It will become 1, 0, 0, 3. Further, I think we will not be able to divide by any number. So, it will become sum of 2003 terms will become 2 into 4003 will be 4006 upon 3 into 1003 will be 3009, right. So, this will be our answer. 4006 upon 3009. So, let's see whether it is there in option or not. 4006 upon 3300 linear. It's there in option D. So, option D is right. So, let's move to the next question, question number 9. The value of 1 upon A 1 plus A 2 plus A into 2 plus A plus 1 upon 2 plus A into 3 plus A and this series is going up to infinity. This series is going up to infinity then we have to find the value of this series where A is the any constant, where A is any constant, okay. So, this series, let us write the Rth term first. Let us write the Rth term for this series. It will become 1 upon R plus A into R plus 1 plus A, right. This will be the Rth term. Let's put R equal to 3 here then it will become 1 upon 3 plus A multiplied by 4 plus A 3 plus A plus 4 by A, okay. The third term is going coming correct means the Rth term written is correct. Now, similar process we will approach this question by VN method where we will define VN as 1 upon R plus 1 plus A. So, our V, sorry R since we are writing in terms of R. So, let's write it as VR. So, VR minus 1 will become 1 upon replace R by R minus 1. So, it will become R plus A. Now, take the difference of these two functions. So, VR minus 1 minus VR is equal to 1 upon R plus 1 and minus of VR that is 1 upon R plus 1 plus A, okay. So, this will become R plus A in the LCM into R plus 1 plus A. This will become R plus 1 plus A minus R minus A. So, minus A plus A minus R plus R. So, eventually we got the Rth term itself. So, VR minus 1 minus VR is equal to Rth term itself, right. This is nothing but TR, no? So, our first term will be V0 minus V1, okay. Second term will be V1 minus V2. Third term will be V2 minus V3 similarly. Our Nth term will be V in minus 1 minus of VN. So, sum it. So, we will get a sum of N terms as V0 minus VN, okay. Sum to N terms will be V0 minus VN and what will be V0? V0 will be put R equal to 0 here. So, it will be 1 upon 1 plus A and what will be VN? Replace R by N, it will be 1 upon N plus 1 plus A, okay. Now, in question the series is going up to infinity. The series is going up to infinity means what? We have to take N as infinity. So, our sum to N terms, let me first write the sum to N terms. It will be V0 means 1 upon 1 plus A minus VN will be 1 upon N plus 1 plus A. Now, we have to take this sum when N is approaching to infinity. So, we have to take the limiting case where N is approaching to infinity, okay. When this is the case, when N approaches infinity, this thing will go to 0. 1 upon infinity, this will go to 0. So, our sum to infinity or sum to N terms when N approaches to infinity becomes 1 upon A plus A, 1 upon 1 plus A. So, option A is correct, okay. So, let's do this 10th question. I think this is the last question of this exercise. So, let's do it. It is saying f of x plus y, okay. fx is a function satisfying f of x plus y is equal to f of x into f of y, right, where x, y belongs to natural number, okay. And f1 is given as 3. f1 is given as 3, okay. f1 is given as 3. And question is asking to find the sum of sigma fx where x is going from 1 to N, okay. This is given actually. This is given as 120. Like a summation of fx where x is going from 1 to N is equal to 120. f of 1 is given as 3. And this is this function f of x plus y is equal to f of x into f of y. So, we have to find the value of N, okay. We have to find the value of N for which this equation holds, okay. So, let's take, let's use this equation. So, f of, we know f of x plus y is equal to f of x into f of y, okay. So, let x is equal to 1 and y is equal to 1. So, it will become f of 1 plus 1 that will be 2 is equal to f of 1 into f of 1. And we know the value of f of 1. What is f of 1? This is 3 into 3. 3 into 3. This is nothing but 3 squared. Now, let's put x is equal to 2 and y is equal to 1. If we put x equal to 2, y equal to 1, it will become f of 3. f of 3 will be equal to f of 2 into f of 1. And we know the value of f of 2 as 3 squared. So, this will be 3 squared into 3. This will become 3q. Similarly, if we put x equal to 3 and y is equal to 1, we will be able to find the value of f of 4. That will be nothing but f of 3 into f of 1. That will be 3 raised to power 4, okay. So, now we have to find the value of n here. Let's open this sigma first. So, it will be f of 1 plus f of 2 plus dot dot dot. It is going up to n. So, plus f of n is given to be 120, okay. So, what is f of 1? f of 1 is 3. So, let's write it as 3. f of 2 is f of 2 is 3 squared. So, it is 3 squared. Similarly, f of 3 will be 3 cubed. So, 3 cubed. So, f of n. What will be the f of n? It will be actually 3 raised to power n. And this is given to be 120, okay. So, what is this? This is a GP. This is a GP with first term S3 and the common ratio. First term S3 and the common ratio is also 3. So, let's write this as sum of this GP as 3 into r raised to power n. What is r? r is 3 r raised to power n minus 1 upon r minus 1. That is 3 minus 1 is equal to 120, okay. So, this will become 3. Let's divide this 120 by 3. It will be 40, right. So, 40 and so 3 raised to power n minus 1. This becomes 2, 2 into 40 or 3 raised to power n minus 1 is equal to 80. That is 3 raised to power n is equal to 81 or 3 raised to power n is equal to 3 raised to power 4. That means n is equal to 4. So, this is our answer. n is equal to 4. So, option a is complete means option a is correct for this question. So, we are done with this exercise. Hope like everyone find this video useful because in this exercise we have done a lot of questions on based on the method of summation vn method. It has, it had a lot of questions on vn method. So, I have seen students struggling with this vn method. So, it's not that much difficult. So, this is all for today. So, we will meet once again. Till then, ta ta goodbye. Take care.