Question

The value of $1^{2}-2^{2}+3^{2}-4^{2}+\cdots+11^{2}$ is equal to

• A. $55$
• B. $66$
• C. $77$
• D. $88$
• E. $99$

Answer 1

Answer: (B)

Given series, $1^{2}-2^{2}+3^{2}-4^{2}+\ldots 11^{2} $
$ =\left(1^{2}+3^{2}+5^{2}+\ldots+11^{2}\right) -\left(2^{2}+4^{2}+\ldots+10^{2}\right) $
$=\left(1^{2}+2^{2}+3^{2}+\ldots 11^{2}\right)-2\left(2^{2}+4^{2}\right. \left.+\ldots+10^{2}\right) $
$=\left(1^{2}+2^{2}+\ldots+11^{2}\right)-2 \cdot 2^{2}\left(1^{2}+2^{2}\right.$
$\left.+\ldots+5^{2}\right)$
$=11 \cdot \frac{(11+1)(22+1)}{6}$
$-8 \cdot \frac{5 \cdot(5+1)(10+1)}{6}$
$=\frac{11 \cdot 12 \cdot 23}{6}- \frac{8 \cdot 5 \cdot 6 \cdot 11}{6} $
$\left[\because \Sigma n^{2}=\frac{n(n+1)(2 n+1)}{6}\right] $
$=506-440$
$=66$