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

If the sum of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \dots + 2 \cdot (n-1)^2 + n^2$ (when $n$ is odd) is to be determined,given that for even $n$,the sum is $\frac{n(n+1)^2}{2}$,find the sum when $n$ is odd.

The sum to infinity of the following series $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots$ is

In the sequence of sets $(1,2,3), (4,5,6), (7,8,9,10), \ldots$,the sum of elements in the $50^{th}$ set is

If $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \ldots \text{ to } n \text{ terms} = n(n+1) f(n) - 3n$,then $f(1) =$

The $n$th term of the series $1+3+7+13+21+\ldots$ is $9901$. The value of $n$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module