Question

The sum of $n$ terms of the following series; $1^3+3^3+5^3+7^3+....$ is

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

Answer 1

The correct option is A $n^2(2n^2-1)$

$1^3+3^3+5^3+7^3+......+n^3$

$=\sum _{r=1}^n(2r-1{)}^3$

$=\sum _{r=1}^n(8r^3-12r^2+6r-1)$

$=8\sum _{r=1}^n{r}^3-12\sum _{r=1}^n{r}^2+6\sum _{r=1}^{n}r-\sum _{r=1}^{n}1$

$=8\cdot \frac{\left[n\right(n+1){]}^2}{4}-12\cdot \frac{n(n+1)(2n+1)}{6}+6\cdot \frac{n(n+1)}{2}-n$, (use summation formulas)

$=2n^2(n+1{)}^2-2n(n+1\left)\right(2n+1)+3n(n+1)-n$

$=2n(n+1)\left[n\right(n+1)-2n-1]+n[3n+3-1]$

$=2n(n+1)(n^2-n-1)+n(3n+2)$

$=2n(n+1)(n^2-n)-2n(n+1)+n(3n+2)$

$=2n(n+1)\left(n\right)(n-1)-2n^2+3n^2$

$=2n^2(n^2-1)+n^2$

$=n^2(2n^2-2+1)=n^2(2n^2-1)$