 Hi and welcome to the session. I am Asha and I am going to help you with the following question that says, find this term to n terms of each of the series in with the solution and let the k term given series which is noted by ak. Then ak will be equal to k into k plus 1 into k plus 2. Since the first term is in the form of 1 into 1 plus 1 into 1 plus 2, second term is in the form of 2 into 2 plus 1 into 2 plus 2. Similarly, the third term is in the form of 3 into 3 plus 1 into 3 plus 2. So, the third term will be in the form of k into k plus 1 into k plus 2 and the nth term up to which we have to find the sum is n into n plus 1 into n plus 2. Take summation on both the side k running from 1 to n since we have to find the sum up to n terms ak is equal to summation k is equal to 1 to n. Inside the packet k into k plus 1 into k plus 2 which is further equal to summation k running from 1 to n, k square plus k into k plus 2 which is further equal to summation k running from 1 to n, k2 plus 3 k square plus 2 k is equal to summation k2 k running from 1 to n plus summation of k running from 1 to n, 3 k square plus summation k running from 1 to n to k. This is further equal to summation k running from 1 to n, k2 plus 3 times summation k running from 1 to n, k square plus 2 times summation k running from 1 to n, k. This is equal to n plus 1 whole square upon 4 plus 3 times of summation of k square k running from 1 to n as n into n plus 1 into 2 n plus 1 upon 6 plus 2 times of summation of a k running from 1 to n as n into n plus 1 upon 2. Now we can n into n plus 1 upon 4 common. Here we have n into n plus 1. From here we get 2 n's of 2 n plus 1 plus 4. This is further equal to n into n plus 1 upon 4 n square plus n plus 4 into plus 2 plus 4. This is further equal to n into n plus 1 upon 4 into n square plus 5n plus 6 is equal to n upon 4 to n plus 1. Now splitting the middle term to factorize this bracket can written as n square plus 3n plus 2n plus 6. This is further equal to n upon 4 into n plus 1. Taking n common from first 2 terms we have n plus 3 and 2 common from last 2 terms we have n plus 3 plus we have n upon 4 into n plus 1 into n plus 3 into n plus 2. Answer answer is 2 into 3 plus 2 into 3 into 4 plus 3 into 4 into 5 plus so on up to n into n plus 1 into n plus 2 is equal to n into n plus 1 into n plus 3 into n plus 2 upon 4. Now we can also write it as n into n plus 1 into n plus 2 into n plus 3 upon 4. This completes the session. Take care and have a good day.