Write the sum of the series 2 + 4 + 6 + 8 + ... + 2n.

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Solution

$S_{n} = 2 + 4 + 6 + 8 + . . . + 2 n$
$= \frac{n}{2} [4 + (n - 1) 2] = \frac{n}{2} [2 + 2 n] = n [n + 1]$

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