 Hi and welcome to the session. I am Asha and I am going to help you with the following question that says find the sum of the following series up to n terms. Let us begin with the solution and the given series is 1 cube upon 1 plus 1 cube plus 2 cube upon 1 plus 3 plus 1 cube plus 2 cube plus 3 cube upon 1 plus 3 plus 5 plus so on. Now from this series it implies that the nth term of the series will look like 1 cube plus 2 cube plus 3 cube plus so on up to n cube and in the denominator we have 1 plus 3 plus 5 plus so on up to 2n minus 1. Now the numerator is the sum of tubes of natural numbers and it is formalized n square into n plus 1 square upon 4 and in the denominator we have an AP series with first term 1 and common difference 2. So it is formalized total number of terms are n. So we have n upon 2 into 2 times of first term plus number of terms minus 1 into the common difference which is 2. So this is further equal to n square into n plus 1 whole square upon 4 into twice upon n into. Now in simplifying we have 2n. Now n square cancels out with n square 2 and 2 and we have n plus 1 whole square upon 4. So the nth term let us note it by a n is equal to 1 upon 4 into n square plus 2n plus 1. So this implies will be equal to 1 upon 4 into k square plus 2k plus 1 and since we have to find the sum up to n terms therefore taking summation on both the sides summation akk running from 1 to n is equal to summation 1 upon 4 into k square plus 2k plus 1k running from 1 to n or we have 1 upon 4 summation k square k running from 1 to n plus 2 upon 4 summation k running from 1 to n k plus 1 upon 4 summation k running from 1 to n 1 which is further equal to 1 upon 4 the summation k square k running from 1 to n is n into n plus 1 into 2n plus 1 upon 6 plus 1 upon 2 summation k k running from 1 to n is n into n plus 1 upon 2 plus 1 upon 4 into n is equal to 1 upon 4 taking common also n taking common we have here n plus 1 into 2n plus 1 upon 6 plus from here we have n plus 1 plus 1. So this is further equal to n upon 4 into here we have 2n square plus 3n plus 1 plus the klcm 6lcm so here we have 6n plus 6 plus 6. So this further implies that summation akk running from 1 to n is n upon 4 into 2n square plus 3n plus 6n is 9n and 6 plus 6 is 12 plus 1 13 upon 6. So this further equal to 1 upon 24 into 2n square plus 9n plus 13 thus the sum of the given series is n upon 24 into 2n square plus 9n plus 13. So this completes the session take care and have a good day.