Quantitative Aptitude Ques 1377

Question: The sum $ 11^{2}+12^{2}+…+21^{2}=? $

Options:

A) 2926

B) 3017

C) 3215

D) 3311

Show Answer

Answer:

Correct Answer: A

Solution:

  • We know that, sum of square of 1st n natural numbers is $ 1^{2}+2^{2}+3^{2}+…+n^{2}=\frac{n(n+1)(2n+1)}{6} $

$ \therefore $ $ 11^{2}+12^{2}+….+21^{2} $ = (Sum of squares of numbers from 1 to 21) $ - $ (Sum of square of first 10 natural numbers) $ =\frac{21(21+1)(42+1)}{6}-\frac{10,(10+1)(20+1)}{6} $ $ =\frac{21\times 22\times 43}{6}-\frac{10\times 11\times 21}{6} $ $ =3311-385=2926 $

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