sum_{k=1}^{2n+1} (-1)^{k-1} cdot k^2 is equal | vedclass

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(B) The given sum is $S = 1^2 - 2^2 + 3^2 - 4^2 + \dots - (2n)^2 + (2n+1)^2$. We can group the terms as: $S = (1^2 - 2^2) + (3^2 - 4^2) + \dots + ((2n

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$(n+1)(2n+1)$

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(B) The given sum is $S = 1^2 - 2^2 + 3^2 - 4^2 + \dots - (2n)^2 + (2n+1)^2$. We can group the terms as: $S = (1^2 - 2^2) + (3^2 - 4^2) + \dots + ((2n-1)^2 - (2n)^2) + (2n+1)^2$. Using the identity $a^2 - b^2 = (a-b)(a+b)$,each pair becomes: $(1-2)(1+2) + (3-4)(3+4) + \dots + ((2n-1)-2n)((2n-1)+2n) + (2n+1)^2$. $S = -1(3) - 1(7) - 1(11) - \dots - 1(4n-1) + (2n+1)^2$. $S = -(3 + 7 + 11 + \dots + (4n-1)) + (2n+1)^2$. The sum inside the bracket is an arithmetic progression with $n$ terms,first term $a=3$,and last term $l=4n-1$. Sum $= \frac{n}{2}(3 + 4n - 1) = \frac{n}{2}(4n+2) = n(2n+1)$. Thus,$S = -n(2n+1) + (2n+1)^2$. $S = (2n+1)(-n + 2n + 1) = (2n+1)(n+1)$.

$\sum_{k=1}^{2n+1} (-1)^{k-1} \cdot k^2$ is equal to

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