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.
On
Use generating functions.
Write your sum as:
$\begin{align*} S_{n + 1} &= \sum_{0 \le k \le n + 1} k^3 (n + 1 - k) \end{align*}$
So this is a convolution. It is the coefficients of the series:
$\begin{align*} S(z) &= \sum_{n \ge 0} S_n z^n \\ &= \left( \sum_{k \ge 0} k^3 z^k \right) \cdot \left( \sum_{k \ge 0} k z^k \right) \\ &= \frac{z (1 + 4 z + z^2)}{(1 - z)^4} \cdot \frac{z}{(1 - z)^2} \\ &= \frac{z^2 (1 + 4 z + z^2)}{(1 - z)^6} \end{align*}$
Extract coefficients, using the generalized binomial theorem:
$\begin{align*} (1 + u)^{-m} &= \sum_{k \ge 0} \binom{-m}{k} u^k \\ &= \sum_{k \ge 0} (-1)^k \binom{k + m - 1}{m - 1} u^k \end{align*}$
giving:
$\begin{align*} S_n &= [z^n] \frac{z^2 (1 + 4 z + z^2)}{(1 - z)^6} \\ &= ([z^{n - 2}] + 4 [z^{n - 3}] + [z^{n - 4}]) (1 - z)^{-6} \\ &= \binom{n - 2 + 6 - 1}{6 - 1} + 4 \binom{n - 3 + 6 - 1}{6 - 1} + \binom{n - 4 + 6 - 1}{6 - 1} \\ &= \frac{3 n^5 - 5 n^3 + 2 n}{60} \end{align*}$
Your sum is:
$\begin{align*} S_{n + 1} &= \frac{3 n^5 + 15 n^4 + 25 n^3 + 15 n^2 + 2 n}{60} \\ &= \frac{n (n + 1) (n + 2) (3 n^2 + 6 n + 1)}{60} \end{align*}$
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.