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. 232
B. 248
C. 464
D. 496

Answer

B

Explanation

We have,
`1^(2) + 2.2^(2) + 3^(2) + 2.4^(2) + 5^(2) + 2.6^(2) +`....
A = sum of first 20 terms
B = sum of first 40 terms
`:. A = 1^(2) + 2.2^(2) + 3^(2) + 2.4^(2) + 5^(2) + 2.6^(2) + ..+ 2.20^(2)`
`A = (1^(2) + 2^(2) + 3^(2) + ....+ 20^(2)) + (2^(2) + 4^(2) + 6^(2) + ...+ 20^(2))`
`A = (1^(2) + 2^(2) + 3^(2) + ....+ 20^(2)) + 4(1^(2) + 2^(2) + 3^(2) + ...+ 10^(2))`
`A = (20 xx 21 xx 41)/(6) + (4 xx 10 xx 11 xx 21)/(6)`
`A = (20 xx 21)/(6) (41 + 22) = (20 xx 41 xx 63)/(6)`
Similarly
`B = (1^(2) + 2^(2) + 3^(2) + ...+ 40^(2)) + 4(1^(2) + 2^(2) + ...+ 20^(2))`
`B = (40xx41 xx 81)/(6) + (4xx 20 xx 21 xx 41)/(6)`
`B = (40 xx 41)/(6) (81 + 42) = (40 xx 41 xx 123)/(6)`
Now, `B - 2A = 100 lamda`
`:. (40 xx 41 xx 123)/(6) - (2 xx 20 xx 21 xx 63)/(6) = 100 lamda`
`rArr (40)/(6) (5043 - 1323) = 100 lamda rArr (40)/(6) xx 3720 = 100 lamda`
`rArr 40 xx 620 = 100 lamda rArr lamda = (40 xx 620)/(100) = 248`