Question

Let$\sum _{r=1}^n{r}^4=f\left(n\right)$, then $\sum _{r=1}^n(2r-1{)}^4$ is equal to

• A. $f\left(2n\right)-16f\left(n\right)$
• B. $f\left(2n\right)+7f\left(n\right)$
• C. $f(2n+1)8f\left(n\right)$
• D. none of these

Answer 1

The correct option is A $f\left(2n\right)-16f\left(n\right)$

Given $\sum _{r=1}^n{r}^4=f\left(n\right)$.
Now, $\sum _{r=1}^n{(2r-1)}^4=\sum _{r=1}^{2n}{r}^4-\sum _{r=1}^n{\left(2r\right)}^4$.......(.writing as sum of $2n$ terms$-$sum of all even terms)
$\Rightarrow \sum _{r=1}^n{(2r-1)}^4=f\left(2n\right)-16\sum _{r=1}^n{r}^4$
$\Rightarrow \sum _{r=1}^n{(2r-1)}^4=f\left(2n\right)-16f\left(n\right)$