A) 464
B) 496
C) 232
D) 248
Correct Answer: D
Solution :
\[A={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+......+{{20}^{2}}\] \[{{2}^{2}}+{{4}^{2}}+....+{{20}^{2}}\] \[\therefore \] \[A=\Sigma {{20}^{2}}+4\cdot \Sigma {{10}^{2}}\] \[\text{II}{{\text{I}}_{g}}B=\Sigma {{40}^{2}}+4\cdot \Sigma {{20}^{2}}\] \[B-2A=\Sigma {{40}^{2}}+2\cdot \Sigma {{20}^{2}}-8.\Sigma {{10}^{2}}\] \[=\frac{40\times 41\times 81}{6}+2\times \frac{20\times 21\times 41}{6}\] \[-8\times \frac{10\times 11\times 21}{6}\] \[=\frac{40}{6}(41\times 81+21\times 41-22\times 21)\] \[=\frac{40}{6}(4100+82-462)\] \[=\frac{40}{6}(3720)=24800\] \[\therefore \lambda =248\]You need to login to perform this action.
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