Solve that:-
$2^{2} + 4^{2} + 6^{2} + . . . . . . . . + 100^{2}$

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Solution

solution -
$= 2^{2} + 4^{2} + 6^{2} + . . . . . . . . . {.100}^{2}$
$= (1 \times 2)^{2} + (2 \times 2)^{2} + (3 \times 2)^{2} + . . . . . . . . . . . (50 \times 2)^{2}$
= $2^{2} [1^{2} + 2^{2} + . . . . . . . . {.50}^{2}]$
= $4 [1^{2} + 2^{2} + . . . {.50}^{2}]$
we know the formula for sum of square starting from 1 to n
= $\frac{n (n + 1) (2 n + 1)}{6}$
= $4 [\frac{50 \times 51 \times 101}{6}]$
= $171700$

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