Question

${11}^2+{12}^2+{13}^2+...{20}^2=$

• A. $2481$
• B. $2483$
• C. $2485$
• D. $2487$

Answer 1

The correct option is C

$2485$

Explanation for the correct option:

Given series is

Let$S={11}^2+{12}^2+....+{20}^2$

This can be written as

$S=(1^2+2^2+....+{20}^2)-(1^2+2^2+....+{10}^2)$

Sum of the Square of the first $n$ Natural Numbers

${\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}{\mathbf{3}}^{\mathbf{2}}\mathbf{+}{\mathbf{4}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{n}}^{\mathbf{2}}\mathbf{=}\mathbf{\sum }{\mathit{n}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{6}\mathbf{}}$

$\Rightarrow S=\frac{20(20+1)(2\times 20+1)}{6}-\frac{10(10+1)(2\times 10+1)}{6}$

$\Rightarrow S=\frac{17220-2310}{6}=\mathbf{}\mathbf{2485}$

Hence, Option ‘C’ is Correct.