<s>[INST] Show by induction on $n$ that: $$1^4 + 2^4 +\cdots+n^4=\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30}$$
I proved true for case $n=1$ , assumed true for $n=k$ , but cannot get things to work out. I tried putting the right side over a common denominator and going from there but I'm sure how much it helps.
 [/INST] First, show that this is true for $n=1$:
$\sum\limits_{k=1}^{1}k^4=\frac{6\cdot1^5+15\cdot1^4+10\cdot1^3-1}{30}$
Second, assume that this is true for $n$:
$\sum\limits_{k=1}^{n}k^4=\frac{6n^5+15n^4+10n^3-n}{30}$
Third, prove that this is true for $n+1$:
$\sum\limits_{k=1}^{n+1}k^4=$
$\color{red}{\sum\limits_{k=1}^{n}k^4}+(n+1)^4=$
$\color{red}{\frac{6n^5+15n^4+10n^3-n}{30}}+(n+1)^4=$
$\frac{6n^5+15n^4+10n^3-n}{30}+n^4+4n^3+6n^2+4n+1=$
$\frac{6n^5+15n^4+10n^3-n}{30}+\frac{30n^4+120n^3+180n^2+120n+30}{30}=$
$\frac{6n^5+45n^4+130n^3+180n^2+119n+30}{30}=$
$\frac{6(n+1)^5+15(n+1)^4+10(n+1)^3-(n+1)}{30}$

Please note that the assumption is used only in the part marked red.
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