Sum Of N Terms From General Term

Summing a polynomial general term proceeds by splitting it into standard power sums, a routine JEE Advanced decomposition. Expand T_k = k^2 + 2k, so the sum is Σk^2 + 2Σk = n(n+1)(2n+1)/6 + 2·n(n+1)/2 = n(n+1)(2n+1)/6 + n(n+1). Factor out n(n+1): n(n+1)[(2n+1)/6 + 1] = n(n+1)[(2n+1+6)/6] = n(n+1)(2n+7)/6. Option n(n+1)(2n+1)/6 omits the linear contribution 2Σk. Option n(n+1)(n+2)/3 mishandles the combination of the two sums. Option n(n+1)(2n+5)/6 arises from adding 4 instead of 6 inside the bracket. Hence the sum is n(n+1)(2n+7)/6. The broader technique is linearity of summation: because the sum operator distributes over addition and pulls out constants, any polynomial general term decomposes into a weighted combination of the elementary power sums, and factoring the common term keeps the answer compact, a habit JEE Advanced rewards. It is also worth noting that the final cubic polynomial in n has degree exactly one higher than the quadratic general term, mirroring how summation behaves like a discrete integral that raises polynomial degree by one. Plausibility check: for n = 2 the direct sum is T_1 + T_2 = 3 + 8 = 11, and for n = 3 it is 11 + T_3 = 11 + 15 = 26, while the formula gives 3·4·13/6 = 156/6 = 26, confirming the closed form at two separate values.