 Hello and welcome to the session. In this session we discuss the following questions which says find the sum to n terms of the series 5 plus 11 plus 19 plus 29 plus 41 and so on. Before moving on to the solution let's discuss one very important theorem which says that if the n-th term of a series is given by tn and this is equal to an cube plus bn square plus cn plus b. So then we have the sum of the n terms of the series given by sn would be equal to a into summation n cube plus b into summation n square plus c into summation n plus b into n. Now next we have summation n is equal to n into n plus 1 and this whole upon 2 then summation n square is given by n into n plus 1 the whole and this whole into 2n plus 1 and this whole upon 6 also summation n cube is equal to n into n plus 1 the whole this whole upon 2 and this whole square which could also be written as summation n whole square. This is the key idea that we use in this question. Let's now move on to the solution. We are supposed to find the sum to n terms of this given series. So first of all we assume let s be equal to 5 plus 11 plus 19 plus 29 plus 41 plus and so on plus tn minus 1 plus tn. Now this tn is the nth term of the series s is the n terms of the series. So we have s equal to 5 plus 11 plus 19 plus 29 plus 41 plus and so on plus tn minus 1 plus tn. That is the sum of n terms of the series also we can write s as 5 plus 11 plus 19 plus 29 plus so on plus tn minus 2 plus tn minus 1 plus tn. That is we have shifted each term to the right inside. So we get this. Now next subtracting we get that is we need to subtract this and this. So we would get s minus s which is 0 is equal to 5 plus 11 minus 5 which is 6 plus 19 minus 11, 8 plus 29 minus 19, 10 plus 12 plus and so on plus tn minus tn minus 1. That is we subtract these two terms. Now this term tn is subtracted from 0. So where we have minus tn we get further 0 is equal to 5 plus 6 plus 8 plus 10 plus 12 plus and so on up to n minus 1 terms. That is this forms an ap minus tn. Now when we shift this tn to the left hand side we get tn is equal to 5 plus 6 plus 8 plus 10 plus 12 plus and so on up to n minus 1 terms. This is an ap. We know that sum of an ap is equal to n upon 2 this whole into 2a plus n minus 1 the whole into b. So using this sum of n terms of an ap we can write the sum of this n minus 1 terms of this ap. So it would be tn is equal to 5 plus. Now n upon 2 and here since we have n minus 1 terms so we write n minus 1 upon 2 and this into 2 into a that is 2 into the first term which is 6 plus n minus 1 or you can say n minus 1 minus 1 which would be n minus 2. So n minus 2 into d which is the common difference and that would be 2 since 8 minus 6 is 2 10 minus 8 is 2 12 minus 10 is 2 or you can say we have tn is equal to 5 plus n minus 1 upon 2 this whole into 12 plus 2n minus 4 that is tn is equal to 5 plus n minus 1 upon 2 into 10 minus 4 is 8 plus 2n that is we have tn is equal to 5 plus n minus 1 upon 2 into taking 2 common here we have 2 into 4 plus n the whole now this 2 2 cancels so we have tn is equal to 5 plus n minus 1 whole into 4 plus n this gives us tn equal to 5 plus n square plus 4n minus n minus 4 this means that tn is equal to n square plus 3n plus 1 so we have got this nth term of the given series and from this theorem or the result stated in the key idea we know that if we have the nth term of the series then we can find out the sum of the n terms of the series by this so this means s which is the sum of n terms of the series is equal to summation n square plus 3 into summation n plus n into 1 that is s is equal to now summation n square is n into n plus 1 the whole into 2n plus 1 the whole and this whole upon 6 so here we have s is equal to n into n plus 1 the whole into 2n plus 1 and this whole upon 6 plus 3 into summation n and summation n is n into n plus 1 the whole and this whole upon 2 so here we have 3 into n into n plus 1 the whole and this whole upon 2 plus n into 1 which is n so here we can take n upon 2 common and so here we have n upon 2 into n plus 1 into 2n plus 1 the whole and this whole upon 3 plus 3 into n plus 1 the whole upon 1 plus 2 this is equal to n upon 2 now taking lcm here the lcm would be 3 so here we have n plus 1 into 2n plus 1 which would be 2n square plus 3n plus 1 plus 3 into 3 is 9 into n plus 1 becomes 9n plus 9 plus 3 into 2 is 6 and this is equal to n upon 2 into 2n square plus 12n plus 16 and this whole upon 3 now taking 2 common in this numerator we get 2 into n square plus 6n plus 8 and this whole upon 3 now here 2 cancels with 2 so this is equal to n upon 3 into n square plus 6n plus 8 the whole this is s that is we have the sum of the n terms of the series is equal to n upon 3 into n square plus 6n plus 8 the whole this is what we were supposed to find this is our final answer this completes the session hope you understood the solution of this question