C-13-Q153mediumsingle_mcq12+22+⋯+202=?1^{2} + 2^{2} + \cdots + 20^{2} = ?12+22+⋯+202=?a287028702870b280028002800c290029002900d29252925ব্যাখ্যাUsing ∑1nk2=n(n+1)(2n+1)6\sum_{1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}∑1nk2= with : .