Question

$\sum _{k=1}^{2n+1}{(-1)}^{k-1}{\left(k\right)}^2$equals

• A. $(n-1)(2n-1)$
• B. $(n+1)(2n+1)$
• C. $(n+1)(2n-1)$
• D. $(n-1)(2n+1)$

Answer 1

The correct option is B

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

Explanation for the correct answer:

Simplifying the given expression:

$$\begin{gathered} \Rightarrow \sum _{k=1}^{2n+1}{(-1)}^{k-1}{\left(k\right)}^2 \\ \Rightarrow 1^2-2^2+3^2-4^2+\dots {\left(2n\right)}^2+{(2n+1)}^2[\because {(-1)}^{k-1}{\left(k\right)}^2=1^2-2^2+3^2-4^2+\dots {\left(2n\right)}^2+{(2n+1)}^2] \\ \Rightarrow \left[1^2+3^2+5^2+...+{(2n+1)}^2\right]-\left[2^2+4^2+6^2+...+{\left(2n\right)}^2\right] \\ \Rightarrow \left[1^2+3^2+5^2+...+{(2n+1)}^2\right]-2\left[2^2+4^2+6^2+...+{\left(2n\right)}^2\right] \\ \Rightarrow \sum {(2n+1)}^2-2\left(2^2\right)[1+2^2+3^3+\dots n^2]\dots \left(1\right) \end{gathered}$$

We know $\sum n^2=\frac{n(n+1)(2n+1)}{6}$

Equation $\left(1\right)$ becomes,

= $$\begin{gathered} \Rightarrow (2n+1)(2n+1+1)\left(2\right(2n+1)+1)-8\left[\frac{n(n+1)(2n+1)}{6}\right] \\ \Rightarrow (2n+1)(2n+1)\left(\frac{2(4n+3)-8n}{6}\right) \\ \Rightarrow (n+1)(2n+1)\left(\frac{8n+6-8n}{6}\right) \\ \Rightarrow (n+1)(2n+1) \end{gathered}$$

Thus, $\sum _{k=1}^{2n+1}{(-1)}^{k-1}{\left(k\right)}^2=(n+1)(2n+1)$

Hence, option (B) is correct.