A) 385
B) 2916
C) 540
D) 384
Correct Answer: D
Solution :
We know that \[{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+......+{{n}^{2}}\] \[=\frac{n(n+1)(2n+1)}{6}\] \[\therefore \] \[{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+.....+{{10}^{2}}\] \[=({{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{10}^{2}})-1\] \[=\frac{10(10+1)(2\times 10+1)}{6}-1\] \[=\frac{10\times 11\times 21}{6}-1\] \[=358-1=384\]You need to login to perform this action.
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