If 1^2+2^2+3^2+\ldots+n^2 = \dfrac{n(n+1)(2n+1)}{6}, find the value… — WBCS Preliminary Examination 2019 Q151

Q151

ArithmeticNumber Series & Progressions

If $1^2+2^2+3^2+\ldots+n^2 = \dfrac{n(n+1)(2n+1)}{6}$, find the value of $5^2+6^2+7^2+\ldots+10^2$.

Answer: (C) 355

$5^2+\ldots+10^2 = (1^2+2^2+\ldots+10^2) - (1^2+2^2+3^2+4^2) = \dfrac{10\times11\times21}{6} - \dfrac{4\times5\times9}{6} = 385 - 30 = 355$.

WBCS Preliminary Examination 2019 — General Studies · Question 151 of 200