Sum of the series 1 cdot 2015 + 2 cdot 2014 + | vedclass

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(D) Let $n = 2015$. The general term of the series is $T_k = k(n - k + 1)$ for $k = 1, 2, \dots, n$.We need to find the sum $S = \sum_{k=1}^{n} k(n -

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(D) Let $n = 2015$. The general term of the series is $T_k = k(n - k + 1)$ for $k = 1, 2, \dots, n$.We need to find the sum $S = \sum_{k=1}^{n} k(n - k + 1) = \sum_{k=1}^{n} (k(n+1) - k^2)$.Using the summation formulas $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ and $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:$S = (n+1) \sum_{k=1}^{n} k - \sum_{k=1}^{n} k^2 = (n+1) \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}$.Factor out $\frac{n(n+1)}{6}$:$S = \frac{n(n+1)}{6} [3(n+1) - (2n+1)] = \frac{n(n+1)}{6} [3n + 3 - 2n - 1] = \frac{n(n+1)(n+2)}{6}$.Substituting $n = 2015$:$S = \frac{2015 	imes 2016 	imes 2017}{6} = 2015 	imes 336 	imes 2017$.Comparing this with the options,none of the given options match the result $2015 	imes 336 	imes 2017$ exactly as written in the choices provided.

Sum of the series $1 \cdot 2015 + 2 \cdot 2014 + 3 \cdot 2013 + \dots + 2015 \cdot 1$ is equal to :-

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