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+2.2^2+3^2+2.4^2+5^2+2.6^2+...` If `B-2A=100lambda` then `lambda` is equal to (1) 232 (2) 248 (3) 464 (4)496

Options

A. 496
B. 232
C. 248
D. 464

Answer

C

Explanation

Let the given series have n terms , where n is even
`therefore S_n =1^2+2 xx 2^2+3^2+2xx 4^2+5^2+2xx 6^2+….`
`=(1^2+2^2+3^2+……+n^2)+(2^2+4^2+6^2+…..+n^2)`
`(n(n+1)(2n+1))/(6)+(4xxn/2xx(n/2+1)(n+1))/(6)`
`(n(n+1))/(6)[2n+1+n+2]`
`=(n(n-1)^2)/(2)`m
Given that B= `S_(40) " and " A=S_(20)`
Also `100 lambda=B-2A`
`(40xx(41)^2)/2-1xx(20xx(21)^2)/2`
`=20 xx 20 xx 62`
` therefore lambda = 248`