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 B (n+1)(2n+1)

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

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

$=\{1^2+3^2+5^2+....(2n+1){\}}^2-(2^2+4^2+6^2+...\left(2n{)}^2\right)+(2^2+4^2+6^2+...(2n{)}^2)-(2^2+4^2+6^2.....\left(2n{)}^2\right)$

$=1^2+2^2+3^2+...(2n+1{)}^2-2[2^2+4^2+6^2....\left(2n{)}^2\right]$

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

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

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