| The sum \[{{11}^{2}}+{{12}^{2}}+...+{{21}^{2}}=?\] |
A) 2926
B) 3017
C) 3215
D) 3311
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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