How do I prove that $1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$? How do I prove that $1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$?
I've spent quite some time on this problem. So far, I've simplified the right-hand side to $\frac{1}{30}(n+1)[(2n+3)(3n^3)+n(n-1)]$. But then, the algebra becomes very complicated when I add $(n+1)^4$ to both sides of the inductive hypothesis.
 A: your formula can be simplified to $$1/30\,n \left( 2\,n+1 \right)  \left( n+1 \right)  \left( 3\,{n}^{2}+3
\,n-1 \right) 
$$
for $n=1$ we have $$1^4=\frac{1}{30}\cdot 3\cdot 2\cdot 5$$ this is right
now we assume that $$1^4+2^4+3^4+...+k^4=\frac{1}{30}k(2k+1)(k+1)(3k^2+3k-1)$$
now we have to prove that
$$1+2^4+3^4+...+k^4+(k+1)^4=\frac{1}{30}(k+1)(2k+3)(k+2)(3(k+1)^2+3(k+1)-1)$$
have you got it?
now you have to Show that
$$(k+1)^4+\frac{1}{30}k(2k+1)(k+1)(3k^2+3k-1)=\frac{1}{30}(k+1)(2k+3)(k+2)(3(k+1)^2+3(k+1)-1)$$
can you finish?
A: The usual trick is to evaluate the sum $$\sum_{k=1}^n k(k+1)(k+2)(k+3)$$ using  $$k(k+1)(k+2)(k+3) = \dfrac{1}{5}k(k+1)(k+2)(k+3)(k+4) - \dfrac{1}{5}(k-1)k(k+1)(k+2)(k+3).$$ 
Then simply expand the summands to obtain $\sum_{k=1}^n k^4$.
A: We have 
\begin{eqnarray*}
1^4+\cdots + n^4 &=& \frac{1}{5} n^5+\frac{1}{2}n^4+\frac{1}{3}n^3+\frac{1}{30}n) \\
&=& \frac{1}{30}(6 n^5+15n^4+10n^3-n). \\
\end{eqnarray*}
So
\begin{eqnarray*}
1^4+\cdots + n^4 +(n+1)^4 &=& \frac{1}{30}(6 n^5+15n^4+10n^3-n +30n^4+120n^3+180n^2+120n+30). \\
\end{eqnarray*}
Now check this really does equal the expression below
\begin{eqnarray*}
 \frac{1}{30}(6 (n^5 +5n^4+10n^3+10n^2+5n+1) \\
 +15(n^4+4n^3+6n^2+4n+1) \\
 +10(n^3+3n^2+3n+1) \\
 -(n+1) ). \\
\end{eqnarray*}
So 
\begin{eqnarray*}
1^4+\cdots + n^4 +(n+1)^4 &=& \frac{1}{30}(6 (n+1)^5+15(n+1)^4+10(n+1)^3-(n+1)). \\
\end{eqnarray*}
A: $$\underbrace{1^4+2^4+3^4\cdots\ + n^4}_{\sum_{k = 1}^n k^4} = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$$
See the Faulhaber's formula and the Bernoulli number: $\text{B}_{j}(x)$
\begin{align}
\sum_{k=0}^{n} k^p = \frac{1}{p+1} \sum_{j=0}^p (-1)^j {p+1 \choose j} B_j n^{p+1-j}=\frac{\text{B}_{p+1}(n+1)-\text{B}_{p+1}(0)}{p+1}
\end{align}
In your case
\begin{align}
\sum_{k = 1}^n k^4 &= \frac{\text{B}_{4+1}(n+1)-\text{B}_{4+1}(0)}{4+1}\\
&= \frac{1}{5}\text{B}_{5}(n+1)\\
&=\frac{1}{5}\left(\frac16 (-1 - n) + \frac53 (1 + n)^3 - \frac52 (1 + n)^4 + (1 + n)^5\right)\\
&=\frac{1}{5}\left(\frac{\left(-1-n\right)}{6}+\frac{5\left(1+n\right)^3\cdot 2}{6}-\frac{5\left(1+n\right)^4\cdot 3}{6}+\frac{\left(1+n\right)^5\cdot 6}{6}\right)\\
&=\frac{1}{5}\left(\frac{6n^5+15n^4+10n^3-n}{6}\right)\\
&=\frac{6n^5+15n^4+10n^3-n}{30}\\
&= \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n \tag*{$\Box$}
\end{align}
