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