5^{2}+6^{2}+7^{2}+ldots+20^{2} = | vedclass

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

Question

(B) We need to find the sum $S = 5^{2} + 6^{2} + 7^{2} + \ldots + 20^{2}$.This can be expressed as the difference of two sums of squares:$S = \sum_{k=

Vedclass · Accepted Answer

$2840$

Vedclass · Answer

(B) We need to find the sum $S = 5^{2} + 6^{2} + 7^{2} + \ldots + 20^{2}$.This can be expressed as the difference of two sums of squares:$S = \sum_{k=1}^{20} k^{2} - \sum_{k=1}^{4} k^{2}$.Using the formula $\sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}$:For $n=20$: $\sum_{k=1}^{20} k^{2} = \frac{20(21)(41)}{6} = 10 	imes 7 	imes 41 = 2870$.For $n=4$: $\sum_{k=1}^{4} k^{2} = \frac{4(5)(9)}{6} = 2 	imes 5 	imes 3 = 30$.Therefore,$S = 2870 - 30 = 2840$.

$5^{2}+6^{2}+7^{2}+\ldots+20^{2} =$

Similar Questions

Let $a_{1}, a_{2}, a_{3}, \ldots$ be an $A.P.$ If $\sum_{r=1}^{\infty} \frac{a_{r}}{2^{r}}=4$,then $4 a_{2}$ is equal to

Find the fractional value of $0.1232323......$

$\sum_{k=1}^5 \frac{1^3+2^3+\ldots+k^3}{1+3+5+\ldots+(2 k-1)}$ is equal to (in $.5$)

For any integer $n \geq 1$,$\sum_{K=1}^n K(K+2) =$

$2 + 4 + 7 + 11 + 16 + \dots$ to $n$ terms =

Vedclass Test Series

Exam Paper Generator

Online Exam Module