$2 + 4 + 6 + . . + 2 n = n (n + 1)$

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Solution

$2 + 4 + 6 + . . . . . . . . . + 2 n 2 [1 + 2 + 3 + . . . . . . . . . + n] = 2 \times (\frac{n (n + 1)}{2}) = n (n + 1)$

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2 + 4 + 6 + . . . . + 2 n = n (n + 1)

2 + 4 + 6 + . . . + 2 n = m^{2} \times n (n + 1)

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