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 $t_r = 1^2 + 3^2 + 5^2 + \dots + (2r - 1)^2$.$t_r = \sum_{k=1}^r (2k - 1)^2 = \sum_{k=1}^r (4k^2 - 4k + 1)$.Using

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The $r$-th term of the series is $t_r = 1^2 + 3^2 + 5^2 + \dots + (2r - 1)^2$.$t_r = \sum_{k=1}^r (2k - 1)^2 = \sum_{k=1}^r (4k^2 - 4k + 1)$.Using standard summation formulas:$t_r = 4 \cdot \frac{r(r+1)(2r+1)}{6} - 4 \cdot \frac{r(r+1)}{2} + r$.Simplifying this expression:$t_r = \frac{2r(r+1)(2r+1)}{3} - 2r(r+1) + r = \frac{2r(2r^2 + 3r + 1) - 6r^2 - 6r + 3r}{3} = \frac{4r^3 + 6r^2 + 2r - 6r^2 - 3r}{3} = \frac{4r^3 - r}{3}$.The sum of $n$ terms is $S_n = \sum_{r=1}^n t_r = \sum_{r=1}^n \frac{4r^3 - r}{3} = \frac{4}{3} \sum_{r=1}^n r^3 - \frac{1}{3} \sum_{r=1}^n r$.$S_n = \frac{4}{3} \left[ \frac{n(n+1)}{2} \right]^2 - \frac{1}{3} \left[ \frac{n(n+1)}{2} \right]$.$S_n = \frac{n(n+1)}{6} [2n(n+1) - 1] = \frac{1}{6} n(n+1)(2n^2 + 2n - 1)$.

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

Similar Questions

The number of solutions of $2^{1+|\cos x|+|\cos x|^2+\ldots} = 4$ in $(-\pi, \pi)$ is

If $\alpha \in R, n \in N$ and $n+2(n-1)+3(n-2)+\ldots+(n-1)2+n.1 = \alpha n(n+1)(n+2)$,then $\alpha =$

For $\frac{2^2+4^2+6^2+\ldots+(2n)^2}{1^2+3^2+5^2+\ldots+(2n-1)^2}$ to exceed $1.01$,the maximum value of $n$ is

The mean of the squares of the first $n$ natural numbers is

The sum,$\sum_{n=1}^{7} \frac{n(n+1)(2n+1)}{4}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module