Question

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

• 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 A $(n+1)(2n+1)$

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

$=1^2-2^2+3^2-4^2+5^2......-(2n{)}^2+(2n+1{)}^2$

$=1^2+3^2+5^2+.....+(2n+1{)}^2-[2^2+4^2+...+\left(2n{)}^2\right]$

$=1^2+2^2+3^2+.....+(2n+1{)}^2-2[2^2+4^2+....\left(2n{)}^2\right]$

Adding and subtract even terms to complete the series.

Using identify $1^2+2^2+....+n^2=\frac{n(n+1)(2n+1)}{6}$

The addition will become

$=1^2+2^2+3^2+....+(2n+1)-2\left(2^2\right)[1^2+2^2+.....+n^2]$

$=\frac{(2n+1)(2n+2)(4n+3)}{6}-\frac{8n(n+1)(2n+1)}{6}$

$=\frac{2(2n+1)(n+1)(4n+3)}{6}-\frac{8n(n+1)(2n+1)}{6}$

$=\frac{2}{6}(n+1)(2n+1)[4n+3-4n]$

$=\frac{1}{3}(n+1)(2n+1)\left(3\right)$

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