The sum to n terms of the series 1^2 + (1^2 + | vedclass

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

Question

(A) The $r$-th term of the series is $t_r = 1^2 + 3^2 + 5^2 + ... + (2r - 1)^2$.Using the formula for the sum of squares of the first $r$ odd numbers,

Vedclass · Accepted Answer

$\frac{1}{6}n(n + 1)(2n^2 + 2n - 1)$

Vedclass · Answer

(A) The $r$-th term of the series is $t_r = 1^2 + 3^2 + 5^2 + ... + (2r - 1)^2$.Using the formula for the sum of squares of the first $r$ odd numbers,$t_r = \sum_{k=1}^r (2k - 1)^2 = \sum_{k=1}^r (4k^2 - 4k + 1)$.$t_r = 4 \sum_{k=1}^r k^2 - 4 \sum_{k=1}^r k + \sum_{k=1}^r 1 = 4 \cdot \frac{r(r+1)(2r+1)}{6} - 4 \cdot \frac{r(r+1)}{2} + r$.$t_r = \frac{2r(r+1)(2r+1)}{3} - 2r(r+1) + r = \frac{r}{3} [2(2r^2 + 3r + 1) - 6(r+1) + 3] = \frac{r}{3} [4r^2 + 6r + 2 - 6r - 6 + 3] = \frac{r(4r^2 - 1)}{3} = \frac{4r^3 - r}{3}$.Now,the sum of $n$ terms is $S_n = \sum_{r=1}^n t_r = \sum_{r=1}^n \frac{4r^3 - r}{3} = \frac{4}{3} \sum_{r=1}^n r^3 - \frac{1}{3} \sum_{r=1}^n r$.$S_n = \frac{4}{3} \cdot \frac{n^2(n+1)^2}{4} - \frac{1}{3} \cdot \frac{n(n+1)}{2} = \frac{n^2(n+1)^2}{3} - \frac{n(n+1)}{6}$.$S_n = \frac{n(n+1)}{6} [2n(n+1) - 1] = \frac{n(n+1)(2n^2 + 2n - 1)}{6}$.

The sum to $n$ terms of the series $1^2 + (1^2 + 3^2) + (1^2 + 3^2 + 5^2) + ...$ is

Similar Questions

The $p^{th}$ term of the series $\left( 3 - \frac{1}{n} \right) + \left( 3 - \frac{2}{n} \right) + \left( 3 - \frac{3}{n} \right) + \dots$ will be

Find the $25^{th}$ common term of the following $A.P.'s$
$S_1 = 1, 6, 11, .....$
$S_2 = 3, 7, 11, .....$

If the sum of first $n$ terms of an $A.P.$ is $cn(n - 1)$,where $c \neq 0$,then the sum of the squares of these terms is

The harmonic mean of $\frac{a}{1 - ab}$ and $\frac{a}{1 + ab}$ is

If the $19^{th}$ term of a non-zero $A.P.$ is zero,then the ratio of its ($49^{th}$ term) to ($29^{th}$ term) is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module