JEE Main On 08 April 2018 Question 29
Question: Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series $ 1^{2}+2\cdot 2^{2}+3^{2}+2\cdot 4^{2}+5^{2}+2\cdot 6^{2}+…… $ . If $ B-2A=100\lambda $ , then $ \lambda $ is equal to: [JEE Main Online 08-04-2018]
Options:
A) 464
B) 496
C) 232
D) 248
Show Answer
Answer:
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} $
$ II{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 $
