The sum of the series 1 · n + 2 · ( n - 1 ) + 3 · ( n - 2 ) + ? + n · 1 is

Question

The sum of the series $1 \cdot n + 2 \cdot (n - 1) + 3 \cdot (n - 2) + \dots + n \cdot 1$ is

Smart Achievers · Accepted Answer

(1)

$\sum_{r = 1}^{n} r (n - r + 1) = \sum_{r = 1}^{n} (n + 1) r - \sum_{r = 1}^{n} r^{2}$

$= (n + 1) \sum n - \sum n^{2} = \frac{{(n + 1)}^{2} n}{2} - \frac{n (n + 1) (2 n + 1)}{6}$

$= \frac{n (n + 1)}{6} (3 n + 3 - 2 n - 1) = \frac{n (n + 1) (n + 2)}{6}$

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