Question

Sum of $n$ terms of the series $1^3+3^3+5^3+7^3+\ldots$. is

• A. $n^2\left(2 n^2-1\right)$
• B. $n^3(n-1)$
• C. $n^3+8 n+4$
• D. $2 n^4+3 n^2$

Answer 1

Answer: (A)

$T_n=(2 n-1)^3$
$=8 n^3-1^3-3 \cdot 2 n \cdot 1(2 n-1)$
$=8 n^3-1-12 n^2+6 n$
$=8 n^3-12 n^2+6 n-1$
$\therefore S_n=\Sigma T_n$
$=8 \Sigma n^3-12 \Sigma n^2+6 \Sigma n-\Sigma 1$
$=8 \cdot\left[\frac{n(n+1)}{2}\right]^2-12 \cdot \frac{n(n+1)(2 n+1)}{6}$ $+6 \frac{n(n+1)}{2}-n$
$=2 n^2(n+1)^2-2 n(n+1)(2 n+1)+3 n(n+1)-n$
$=n(n+1)[2 n(n+1)-2(2 n+1)+3]-n$
$=n(n+1)\left[2 n^2+2 n-4 n-2+3\right]-n$
$=n(n+1)\left[2 n^2-2 n+1\right]-n$
$=n(n+1) \cdot 2 n(n-1)+n(n+1)-n$
$=2 n^2\left(n^2-1\right)+n^2$
$=n^2\left(2 n^2-1\right)$