Question

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

• A. $f\left(2n\right)-16f\left(n\right)$ for all $nϵN$
• B. $f\left(n\right)-16f\left(\frac{n-1}{2}\right)$ when $n$ is odd
• C. $f\left(n\right)-16f\left(\frac{n}{2}\right)$ when $n$ is even
• D. none of these

Answer 1

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

Given $f\left(n\right)={\sum }_{r=1}^n{r}^4=1^4+2^4+3^4+..........{\left(n-1\right)}^4+n^4$
Now ${\sum }_{r=1}^n{(2r-1)}^4=1^4+3^4+5^4+......+{(2n-1)}^4=1^4+2^4+3^4+..........{\left(n-1\right)}^4+n^4+{\left(n+1\right)}^4+......+{\left(2n\right)}^4-[2^4+4^4+6^4+........+{\left(2n-2\right)}^4+{\left(2n\right)}^4]$
Here we can see that the first 2n terms are just $f\left(2n\right)$
and if we take $2^4$ common from the second n-terms of the expression then it reduces to
$f\left(2n\right)-2^4[1^4+2^4+3^4+.........+n^4]=f\left(2n\right)-16f\left(n\right)$