11^2 + 12^2 + 13^2 + dots + 20^2 = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The sum of the squares of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.To find the sum $11

Vedclass · Accepted Answer

$2485$

Vedclass · Answer

(C) The sum of the squares of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.To find the sum $11^2 + 12^2 + \dots + 20^2$, we calculate the sum of the first $20$ squares and subtract the sum of the first $10$ squares.Sum of first $20$ squares: $\sum_{k=1}^{20} k^2 = \frac{20(20+1)(2 	imes 20 + 1)}{6} = \frac{20 	imes 21 	imes 41}{6} = 10 	imes 7 	imes 41 = 2870$.Sum of first $10$ squares: $\sum_{k=1}^{10} k^2 = \frac{10(10+1)(2 	imes 10 + 1)}{6} = \frac{10 	imes 11 	imes 21}{6} = 5 	imes 11 	imes 7 = 385$.Required sum $= 2870 - 385 = 2485$.

$11^2 + 12^2 + 13^2 + \dots + 20^2 = $

Similar Questions

The sum of the series $(2)^2 + 2(4)^2 + 3(6)^2 + ...$ up to $10$ terms is:

The sum of the first $n$ terms of the series $\cot^{-1} 3 + \cot^{-1} 7 + \cot^{-1} 13 + \cot^{-1} 21 + \dots$ is given by:

If the $7^{th}$ term of a harmonic progression is $8$ and the $8^{th}$ term is $7$,then its $15^{th}$ term is

The value of $1^2 + 3^2 + 5^2 + ....... + 25^2$ is

The value of $\frac{1}{(1 + a)(2 + a)} + \frac{1}{(2 + a)(3 + a)} + \frac{1}{(3 + a)(4 + a)} + \dots + \infty$ is,(where $a$ is a constant).

Vedclass Test Series

Exam Paper Generator

Online Exam Module