If the sum of the first n terms of an arithme | vedclass

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

Question

(C) Given $S_n = cn^2$.$S_{n-1} = c(n-1)^2$.The $n^{th}$ term $t_n = S_n - S_{n-1} = cn^2 - c(n-1)^2 = c(2n-1)$.The sum of the squares of these $n$ te

Vedclass · Accepted Answer

$\frac{n(4n^2 - 1)c^2}{3}$

Vedclass · Answer

(C) Given $S_n = cn^2$.$S_{n-1} = c(n-1)^2$.The $n^{th}$ term $t_n = S_n - S_{n-1} = cn^2 - c(n-1)^2 = c(2n-1)$.The sum of the squares of these $n$ terms is $\sum_{k=1}^{n} t_k^2 = \sum_{k=1}^{n} c^2(2k-1)^2$.$= c^2 \sum_{k=1}^{n} (4k^2 - 4k + 1)$.$= c^2 [4 \sum k^2 - 4 \sum k + \sum 1]$.$= c^2 [4 \frac{n(n+1)(2n+1)}{6} - 4 \frac{n(n+1)}{2} + n]$.$= c^2 [\frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n]$.$= \frac{c^2 n}{3} [2(2n^2 + 3n + 1) - 6(n+1) + 3]$.$= \frac{c^2 n}{3} [4n^2 + 6n + 2 - 6n - 6 + 3]$.$= \frac{n(4n^2 - 1)c^2}{3}$.

If the sum of the first $n$ terms of an arithmetic progression is $cn^2$,what is the sum of the squares of these $n$ terms?

Similar Questions

The number of terms in the series $101 + 99 + 97 + \dots + 47$ is

If $< a_n >$ is an $A.P.$ and $a_1 + a_4 + a_7 + \dots + a_{16} = 147$,then $a_1 + a_6 + a_{11} + a_{16}$ is equal to

$150$ workers were engaged to finish a piece of work in a certain number of days. $4$ workers dropped the second day,$4$ more workers dropped the third day and so on. It takes eight more days to finish the work now. The number of days in which the work was completed is

If the $7^{th}$ term of an Arithmetic Progression $(AP)$ is $40$,then the sum of its first $13$ terms is........

If the $n^{th}$ term of an arithmetic progression is $\frac{(2n + 1)}{3}$,what is the sum of its first $19$ terms?

Vedclass Test Series

Exam Paper Generator

Online Exam Module