C-13-Q149mediumsingle_mcq82+92+102+⋯+1528^{2} + 9^{2} + 10^{2} + \cdots + 15^{2}82+92+102+⋯+152 equals —a110011001100b124012401240c138013801380d145014501450ব্যাখ্যাUse ∑1nk2=n(n+1)(2n+1)6\sum_{1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}∑1nk2=6n(n+1)(2n+1). Then ∑115k2=1240\sum_{1}^{15}k^2=1240∑115k2=1240 and ∑17k2=140\sum_{1}^{7}k^2=140∑17k2=140, so the required sum is 1240−140=11001240-140=11001240−140=1100.