Let A be the sum of the first 20 terms and B | vedclass

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(A) The series is $a_n = n^2$ if $n$ is odd,and $a_n = 2n^2$ if $n$ is even.$B - 2A = \sum_{n=1}^{40} a_n - 2\sum_{n=1}^{20} a_n = \sum_{n=21}^{40} a_

Vedclass · Accepted Answer

$248$

Vedclass · Answer

(A) The series is $a_n = n^2$ if $n$ is odd,and $a_n = 2n^2$ if $n$ is even.$B - 2A = \sum_{n=1}^{40} a_n - 2\sum_{n=1}^{20} a_n = \sum_{n=21}^{40} a_n - \sum_{n=1}^{20} a_n$.For $n \in [21, 40]$,let $k = n-20$,so $n = k+20$. The terms are $(k+20)^2$ if $k$ is odd,and $2(k+20)^2$ if $k$ is even.$B - 2A = \sum_{k=1}^{20} (a_{k+20} - a_k)$.If $k$ is odd,$a_{k+20} - a_k = (k+20)^2 - k^2 = 40k + 400$.If $k$ is even,$a_{k+20} - a_k = 2(k+20)^2 - 2k^2 = 2(40k + 400) = 80k + 800$.Summing these for $k=1$ to $20$ ($10$ odd and $10$ even values of $k$):$B - 2A = \sum_{k \in 	ext{odd}} (40k + 400) + \sum_{k \in 	ext{even}} (80k + 800)$.$= 40 \sum_{k \in 	ext{odd}} k + 4000 + 80 \sum_{k \in 	ext{even}} k + 8000$.$= 40(1+3+\dots+19) + 80(2+4+\dots+20) + 12000$.$= 40(100) + 80(110) + 12000 = 4000 + 8800 + 12000 = 24800$.Since $B - 2A = 100\lambda$,$100\lambda = 24800$,so $\lambda = 248$.

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

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