 Hello and welcome to the session. In this session we will discuss how to determine an explicit expression, a recursion, and steps for calculation from context and combine standard function types using arithmetic operations. For this, let us see an illustration. In this illustration, we will determine the explicit expression as well as the recursion process. And the illustration is, suppose there are 3000 fish currently in the pond, 1000 fish are abit at the end of each year, the population of fish decreases each year by 20%. Now first we will derive recursion process for this illustration. Now in recursion, we find each term in the sequence by using the previous terms when the first term or initial term or initial value is known. Now we know that initial population of fish is equal to 3000 that is P0 is equal to 3000. Now that population of fish after one year is equal to P1. Now we know that 1000 fish are abit at the end of each year and the population of fish decreases each year by 20%. So using these conditions, we will find the population of fish after one year. Now since after one year the population 3000 is decreased by 20%, so after one year remaining population will be 3000 minus 20% of 3000. Also 1000 fish are added to the remaining population, so population after one year that is P1 is equal to 3000 minus 20% of 3000 the whole plus 1000. So this is equal to 3000 minus 20% upon 100 into 3000 the whole plus 1000. Now from these two terms taking 3000 common, so it will be 3000 into 1 minus 20% upon 100 the whole plus 1000. And this is equal to 3000 into 1 minus 0.2 the whole plus 1000 which is equal to 3000 into 0.8 the whole plus 1000. And this is equal to 3000 into 0.8 is 2400 plus 1000 which is equal to 3400. Therefore population of fish after one year that is P1 is equal to 3400. So now population is 3400 in similar way the population after will be P2 which is equal to 3400 into 0.8 plus 1000 which is equal to 2720 plus 1000. That is equal to 3720 now population is 3720 so population next year that is the population after three years will be P3 which is equal to 3720 into 0.8 plus 1000 that is equal to 2976 plus 1000 which is equal to 3976. So we have the following populations year after year P0 is equal to 3000 then P1 is equal to 3000 into 0.8 plus 1000 which is equal to 3400. This can also be written as P1 is equal to P0 that is 3000 into 0.8 plus 1000 now P2 is equal to 3400 into 0.8 plus 1000 which is equal to 3720. And this can be written as P2 is equal to now where you can see P1 is equal to 3400. So here we can write P1 into 0.8 plus 1000 also we have P3 is equal to 3720 into 0.8 plus 1000 which is equal to 3976. So here we can write it as P3 is equal to P2 into 0.8 plus 1000 now continuing like this after 10 years we have population Pn is equal to Pn minus 1 into 0.8 plus 1000 where Pn minus 1 previous year population so we have got recursive formula P0 is equal to 3000 Pn is equal to Pn minus 1 into 0.8 plus 1000. Now here we will derive the explicit formula for this illustration. Now explicit expression or explicit formula answers how to calculate any term of the sequence without calculating previous terms. Now here we have current population of fish which is equal to 3000 and we have given this population decreases each year by 20%. So here we have given decay rate that is r is equal to 20% so decreasing factor of population will be b that is equal to 1 minus r the whole. That is b is equal to 1 minus 20% the whole which is equal to 1 minus 20 upon 100 the whole which is equal to 1 minus 0.2 the whole that is equal to 0.8. So b is equal to 0.8 now let T be the number of years then population P after T years will be an exponential function given by P is equal to a into v raised to power T plus k of P is equal to a into now we have found the value of b that is 0.8 so P is equal to a into 0.8 raised to power T plus k. Now we have given that population decreases by 20% each year and later 1000 is added every year here we cannot directly take initial population to be 3000 in the equation. Now to find values of a and k we will make use of initial conditions by 1 P is equal to 0 the population P is equal to 3000. Now here let us take this as equation number 1 now let this be condition number 1 so when T is equal to 0 population P is equal to 3000. Now the next year population will be 3400 as we have shown it earlier while deriving the recursive relation so when T is equal to 1 the population P is equal to 3400 now let this be condition 2. Now using condition 1 and condition 2 we will find values of a and k from equation 1 now this is equation number 1 so let us put T is equal to 0 and P is equal to 3000 in equation 1 so we get 3000 is equal to a into 0.8 raised to power 0 plus k. Now this implies 3000 is equal to into now 0.8 raised to power 0 is 1 so a into 1 is a plus k now let this be equation number 2 now let us put T is equal to 1 and P is equal to 3400 in equation 1 400 is equal to a into 0.8 raised to power 1 plus k now this implies 3400 is equal to 0.8 into a plus k now let this be equation number 3 now solving equation number 2 and 3 we get equal to minus 2000 and k is equal to 5000 now we will put these values in equation number 1 so putting is equal to minus 2000 k is equal to 5000 in equation 1 we get P is equal to minus 2000 into 0.8 raised to power T plus 5000 So using this formula you can find the population of fish in the ground after any years now suppose we want to find population after formulas simply we put T is equal to 4 in this formula and we have T is equal to minus 2000 into 0.8 raised to power 4 plus 5000 so this is equal to minus 2000 into 0.4096 5000 now on solving this is equal to minus 819.2 plus 5000 which is equal to 4118.8 which is approximately equal to 4181 so the population of fish in plant after 4 years is equal to 4181 now in the above illustration we can see that we have combined an exponential decay function with a constant function given by k using arithmetic operation so in this session we have discussed all the measurement and explicit expression, a recursion process, operation from context and combined using arithmetic operations and this completes our session hope you all have enjoyed the session