If S_n is the sum of the first n terms of the | vedclass

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

Question

(C) The series is $1^2, 2(2^2), 3^2, 2(4^2), 5^2, 2(6^2), \ldots$ For even $n$,let $n = 2m$. The sum $S_{2m}$ consists of $m$ odd-indexed terms and $m

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The series is $1^2, 2(2^2), 3^2, 2(4^2), 5^2, 2(6^2), \ldots$ For even $n$,let $n = 2m$. The sum $S_{2m}$ consists of $m$ odd-indexed terms and $m$ even-indexed terms. $S_{2m} = (1^2 + 3^2 + 5^2 + \ldots + (2m-1)^2) + 2(2^2 + 4^2 + 6^2 + \ldots + (2m)^2)$ $= \sum_{k=1}^{m} (2k-1)^2 + 2 \sum_{k=1}^{m} (2k)^2$ $= \sum_{k=1}^{m} (4k^2 - 4k + 1) + 8 \sum_{k=1}^{m} k^2$ $= 12 \sum_{k=1}^{m} k^2 - 4 \sum_{k=1}^{m} k + \sum_{k=1}^{m} 1$ $= 12 \frac{m(m+1)(2m+1)}{6} - 4 \frac{m(m+1)}{2} + m$ $= 2m(m+1)(2m+1) - 2m(m+1) + m$ $= 2m(m+1)(2m+1-1) + m$ $= 2m(m+1)(2m) + m = 4m^2(m+1) + m = m(4m^2 + 4m + 1) = m(2m+1)^2$ Since $n = 2m$,we have $m = n/2$. Substituting $m = n/2$ into the expression: $S_n = \frac{n}{2}(2(\frac{n}{2}) + 1)^2 = \frac{n}{2}(n+1)^2$.

If $S_n$ is the sum of the first $n$ terms of the series $1^2+2 \times 2^2+3^2+2 \times 4^2+5^2+2 \times 6^2+\ldots$,then for even $n$,$S_n$ is equal to:

Similar Questions

The value of $\sum\limits_{r = 16}^{30} {(r + 2)(r - 3)}$ is equal to

If $3 + 3\alpha + 3\alpha^2 + \dots \infty = \frac{45}{8}$,then the value of $\alpha$ will be

The number of positive integers $n$ in the set $\{1, 2, 3, \ldots, 100\}$ for which the number $\frac{1^2+2^2+3^2+\ldots+n^2}{1+2+3+\ldots+n}$ is an integer is

$\sum_{k=1}^{2n+1} (-1)^{k-1} \cdot k^2$ is equal to

The sum of the following series $1 + 6 + \frac{9(1^2 + 2^2 + 3^2)}{7} + \frac{12(1^2 + 2^2 + 3^2 + 4^2)}{9} + \frac{15(1^2 + 2^2 + ... + 5^2)}{11} + ...$ up to $15$ terms is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module