<s>[INST] Finding a formula for $(1^3)\cdot(n)+(2^3)\cdot(n-1)+(3^3)\cdot(n-2)+ \cdots + (n^3)\cdot(1)$ 
I need a formula for this sum:
  $$(1^3)\cdot(n)+(2^3)\cdot(n-1)+(3^3)\cdot(n-2)+ \cdots + (n^3)\cdot(1)$$

I have found this formula :
$$1\cdot n +2\cdot(n-1)+3\cdot(n-2)+ \cdots +(n-1)\cdot 2 +n\cdot1= \frac16n(n+1)(n+2)$$
but it is not exactly what I need.
Thanks in advance.
 [/INST] I can rewrite the sum as $S_n=\sum\limits _{k=1}^{n}k^3(n+1-k)=(n+1)\sum\limits _{k=1}^{n}k^3 - \sum\limits _{k=1}^{n}k^4$. Now, we can show by induction that $A_n=\sum\limits _{k=1}^{n}k^3=\frac{n^4+2n^3+n^2}{4}$ and $B_n=\sum\limits _{k=1}^{n}k^4=\frac{6n^5+15n^4+10n^3-n}{30}$. Substituing this formulas in: $S_n=(n+1)A_n - B_n$, I have: $$S_n=(n+1)\cdot \frac{n^4+2n^3+n^2}{4}-\frac{6n^5+15n^4+10n^3-n}{30}=\frac{3n^5+15n^4+25n^3+15n^2+2n}{60}$$
This is the expansion of your sum.
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