Correct Answer - Option 3 : 2485
Concept:
Sum of the first n natural numbers
\(\rm 1+2+3+4+....+n =\sum n = \frac{n(n+1)}{2}\)
Sum of the Square of the first n Natural Numbers \(\rm 1^2+2^2+3^2+4^2+....+ n^2=\sum n^2 = \frac{n(n+1)(2n+1)}{6}\)
Sum of the Cubes of the first n Natural Numbers \(\rm 1^3+2^3+3^3+4^3+....+ n^3=\sum n^3 = \frac{[n(n+1)]^2}{4}\)
Calculation:
Given series
= 112 + 122 + ....+ 202
This can be written as
(12 + 22 + ....+ 202) - (12 + 22 + ....+ 102)
∵ \(\rm 1^2+2^2+3^2+4^2+....+ n^2=\frac{n(n+1)(2n+1)}{6}\)
⇒ \(\frac{20(20\ +\ 1)(2\times 20\ +\ 1)}{6}\ -\ \frac{10(10\ +\ 1)(2\times 10\ +\ 1)}{6}\)
⇒ \(\frac{17220\ -\ 2310}{6}\ =\ 2485\)
