Question

Sum of series 2^2+4^2+6^2+.....n^2

Answer 1

$\large \textit{Using the formula}\ \ 1^2+2^2+3^2+...+n^2 \ = \ \displaystyle \frac{n(n+1)(2n+1)}{6}$

$\large \textit{Taking \ n=2k ,} \\ \\ \\ \displaystyle 2^2+4^2+6^2+... 2k^2 \\ \\ \\ 2(1^2+2^2+3^2+...+k^2) \\ \\ \\ 2\left(\frac{k(k+1)(2k+1)}{6}\right) \\ \\ \\ \frac{2k(k+1)(2k+1)}{6} \\ \\ \\ \frac{2 \cdot 2k(k+1)(2k+1)}{2 \cdot 6} \\ \\ \\ \frac{2k \cdot 2(k+1)(2k+1)}{12} \\ \\ \\ \frac{2k(2k+2)(2k+1)}{12} \\ \\ \\ \frac{n(n+1)(n+2)}{12}$

$\large \textit{Thus,} \ \ \ \displaystyle 2^2+4^2+6^2+...+n^2 \ = \ \frac{n(n+1)(n+2)}{12}$