<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] $$\sum_{k=1}^{n+1}k^{4}=\sum_{k=1}^{n}k^{4}+\left(n+1\right)^{4}=\frac{n^{5}}{5}+\frac{n^{4}}{2}+\frac{n^{3}}{3}-\frac{n}{30}+n^{4}+4n^{3}+6n^{2}+4n+1
 $$$$=\frac{6n^{5}+15n^{4}+10n^{3}-n}{30}+\frac{30n^{4}+120n^{3}+180n^{2}+120n+30}{30}=\frac{6n^{5}+45n^{4}+130n^{3}+180n^{2}+119n+30}{30}$$
 and see that$$6\left(n+1\right)^{5}+15\left(n+1\right)^{4}+10\left(n+1\right)^{3}-\left(n+1\right)$$$$=\left(6n^{5}+30n^{4}+60n^{3}+60n^{2}+30n+6\right)
 +\left(15n^{4}+60n^{3}+90n^{2}+60n+15\right)+\left(10n^{3}+30n^{2}+30n+10\right)-n-1=6n^{5}+45n^{4}+130n^{3}+180n^{2}+119n+30.$$
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