Suppose for each $n\in \mathbf{N},1^4+2^4+3^4+\dots +n^4=a{n}^5+b{n}^4+c{n}^3+d{n}^2+en+f$, then value of $b$ is 

We have 
$1^4+2^4+\dots +n^4=a{n}^5+b{n}^4+c{n}^3+d{n}^2+en+f \left(1\right)$
and $1^4+2^4+\dots +n^4+(n+1{)}^4$
       $\begin{array}{r}=a(n+1{)}^5+b(n+1{)}^4+c(n+1{)}^3+\\ d(n+1{)}^2+e(n+1)+f (2)\end{array}$
Subtracting $\left(1\right)$ from $\left(2\right)$, we get
$\begin{array}{l}(n+1{)}^4=a\left[(n+1{)}^5-n^5\right]+b\left[(n+1{)}^4-n^4\right]\\ +c\left[(n+1{)}^3-n^3\right]+d\left[(n+1{)}^2-n^2\right]\\ +e\left[\right(n+1)-n]\end{array}$
Comparing coefficients fo $n^4$ , we get
            $1=5a\Rightarrow a=1/5$
Comparing coefficient fo $n^3$ , we get
          $4=a\left({ }^5{C}_2\right)+4b\Rightarrow b=1/2.$