1^2+left(1^2+2^2right)+left(1^2+2^2+3^2right) | vedclass

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(C) The $n$-th term of the series is $T_n = \sum_{j=1}^n j^2 = \frac{n(n+1)(2n+1)}{6}$.We need to find the sum $S_n = \sum_{i=1}^n T_i = \sum_{i=1}^n

Vedclass · Accepted Answer

$\frac{n(n+1)^2(n+2)}{12}$

Vedclass · Answer

(C) The $n$-th term of the series is $T_n = \sum_{j=1}^n j^2 = \frac{n(n+1)(2n+1)}{6}$.We need to find the sum $S_n = \sum_{i=1}^n T_i = \sum_{i=1}^n \frac{i(i+1)(2i+1)}{6}$.Expanding the expression: $S_n = \frac{1}{6} \sum_{i=1}^n (2i^3 + 3i^2 + i) = \frac{1}{3} \sum i^3 + \frac{1}{2} \sum i^2 + \frac{1}{6} \sum i$.Using standard summation formulas:$S_n = \frac{1}{3} \left[\frac{n(n+1)}{2}\right]^2 + \frac{1}{2} \left[\frac{n(n+1)(2n+1)}{6}\right] + \frac{1}{6} \left[\frac{n(n+1)}{2}\right]$.Factoring out $\frac{n(n+1)}{12}$:$S_n = \frac{n(n+1)}{12} \left[ n(n+1) + (2n+1) + 1 \right] = \frac{n(n+1)}{12} [n^2 + n + 2n + 2] = \frac{n(n+1)(n^2+3n+2)}{12}$.Since $n^2+3n+2 = (n+1)(n+2)$,we get $S_n = \frac{n(n+1)^2(n+2)}{12}$.

$1^2+\left(1^2+2^2\right)+\left(1^2+2^2+3^2\right)+\ldots+\left(1^2+2^2+\ldots+n^2\right)=$

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