Question

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

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

Answer 1

Answer: (A)

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