A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 12+2⋅22+32+2⋅42+52+2⋅62+….…If B−2A=100λ, then λ is equal to

A
496
B
232
C
248
D
464

Solution:

Let $S = 1^{2} + 2 \cdot 2^{2} + 3^{2} + 2 \cdot 4^{2} + 5^{2} + 2 \cdot 6^{2} + \dots$

The sum of first 20 terms is

$\begin{array}{l} A = (1^{2} + 2^{2} + \dots . + 20^{2}) + (2^{2} + 4^{2} + \dots . + 20^{2}) \\ = \frac{20 \cdot 21 \cdot 41}{6} + 4 \cdot \frac{10 \cdot 11 \cdot 21}{6} = \frac{20 \cdot 21}{6} (41 + 22) = 4410 \\ B = 1^{2} + 2 \cdot 2^{2} + \dots . + 2 \cdot 40^{2} \\ = (1^{2} + 2^{2} + \dots . + 40^{2}) + (2^{2} + 4^{2} + \dots . + 40^{2}) \\ = \frac{40 \cdot 41 \cdot 81}{6} + \frac{4 \cdot 20 \cdot 21 \cdot 41}{6} = 33620 \\ B - 2 A = 33620 - 8820 = 24800 \\ ∴ 100 λ = 24800 \Rightarrow λ = 248 \end{array}$

$A$ be the sum of the first 20 terms and $B$ be the sum of the first 40 terms of the series $1^{2} + 2 \cdot 2^{2} + 3^{2} + 2 \cdot 4^{2} + 5^{2} + 2 \cdot 6^{2} + \dots .$ …If $B - 2 A = 100 λ,$ then $λ$ is equal to

Solution:

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