<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] 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:
$\begin{align*}
  S_{n + 1}
    &= \frac{3 n^5 + 15 n^4 + 25 n^3 + 15 n^2 + 2 n}{60} \\
    &= \frac{n (n + 1) (n + 2) (3 n^2 + 6 n + 1)}{60}
\end{align*}$
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