 Hi and welcome to the session. I am Arsha and I am going to help you with the following question which says, find the sum of the first n terms of the series 3 plus 7 plus 13 plus 21 plus 31 plus 2 own. Let us now start with the solution and let us write the given sum of first n terms as Sn. So Sn is equal to 3 plus 7 plus 13 plus 21 plus 31 plus so on up to n minus 1 plus a n or we can also write Sn as 3 plus 7 plus 13 plus 21 plus 31 plus a n minus 2 plus a n minus 1 plus a n. Now in subtracting the first line we have on the left hand side 0 and on the right hand side 3 plus 4 plus 6 minus 1 terms minus a n or this would imply that a n is equal to 3 plus 4 plus 6 plus 8 plus 10 plus so on up to n minus 1 terms. Now as we can see this is an A B series whose first term is 4 and common difference is 2. So the sum of these n minus 1 terms is equal to n minus 1 upon 2 into 2 lines of first term which is 4 plus n minus 1 and here in place of n we have n minus 1. So n minus 2 into d which is 2 that is the common difference is 2. So this is further equal to 3 n minus 1 upon 2 into 8 plus n minus 4 which further implies that 3 plus n minus 1 upon 2 into 4 plus 2 n which further implies that 3 plus n minus 1 upon 2 into 2 lines n plus 2. So this gives us 3 plus n square plus 2 n minus n minus 2 or we have n square plus 1 that is noted by a k will be equal to k square plus k plus 1 and we have to find the sum up to n terms. So taking summation on both the sides we have summation a k running from 1 to n is equal to summation k square plus k plus 1 k running from 1 to n. So this is equal to the sum of n terms we have to find out and the right hand side will be equal to summation k square k running from 1 to n plus summation k k running from 1 to n plus summation 1 k running from 1 to n. So this is further equal to n into n plus 1 into 2 n plus 1 upon 6 plus n into n plus 1 upon 2 plus n. This is further equal to taking n common. Here we have 8 plus 1 into 2 n plus 1 upon 6 plus n plus 1 upon 2 plus 1. This is further equal to n or the LCM is 6 and the numerator we have 2 n square plus 3 n plus 1 into 3 is 6 so we have 3 n which is further equal to n into 2 n square plus 6 n plus 10 upon 6. Now taking 2 common from the numerator we have upon 6 into n square plus 3 n plus 5 this is further equal to n upon 3 into n square plus 3 n plus 5 and hence the answer is the required sum is n upon 3 into n square plus 3 n plus 5. So this completes the session. Take care and have a good day.