 Hi and welcome to the session. I am Asha and I am going to help you with the following question which says, find the sum 2n terms of the series whose nth term is given by n into n plus 1 into n plus 4. So let us begin with the solution. Now we are given at the nth term of the series is equal to n into n plus 1 into n plus 4. Therefore kth term of the series will be k into k plus 1 into k plus 4. So let us denote the kth term by ak. So ak is equal to k into k plus 1 into k plus 4. Now we have to find the sum 2n terms. Therefore taking summation on both the sides, summation ak k running from 1 to n and on the right hand side we have summation k square plus k into k plus 4 summation k running from 1 to n which is further equal to summation k running from 1 to n kq plus 4k square plus k square plus 4k which is further equal to summation k running from 1 to n kq plus 5k square plus 4k which is further equal to summation k running from 1 to n kq plus 5 times summation k running from 1 to n k square plus 4 times summation k k running from 1 to n. So this is further equal to n square into n plus 1 whole square upon 4 plus 5 times of n into n plus 1 into 2n plus 1 upon 6 plus 4 times of summation k k running from 1 to n as n into n plus 1 upon 2. Now taking n into n plus 1 upon 12 common, here we have 3 times of n into n plus 1 plus 10 times of 2n plus 1 plus, now multiplying the numerator and denominator by 6 gives 12 and the denominator. So in the numerator we are left with 12. So this is further equal to n into n plus 1 upon 12 3n square plus 3n plus 20n plus 10 plus 24 which is equal to n into n plus 1 upon 12 3n square plus 23n plus 34. Therefore the sum to n terms or any term is given as n into n plus 1 upon 12 into 3n square plus 23n plus 34. So this completes the session. Take care and have a good day.