Can anyone please provide a simplified formula for the sum of the sequence \begin{equation*} s(n) = 1\cdot n + 2(n-1) + \ldots + (n-1)2 + n\cdot1 \end{equation*} where $n$ is an integer greater than $0$?
for instance, $s(1)=1,~ s(2)=4,~ s(3)=10,~ s(7)=84$ (I believe) but what is, say, $s(15)$? (without resorting to a computer)
Thanks! :)
Hint: Rewrite your sum using sigma notation $$\sum_{i=1}^{n}i(n-(i-1)) $$ Then use basic formulas like: $$\sum_{i=1}^{n}i = \frac{(n)(n+1)}{2} $$ $$\sum_{i=1}^{n}i^2 = \frac{(n)(n+1)(2n+1)}{6} $$ to simplify your expression.
$$\sum_{k=0}^{n+1} k(n+1-k)=\sum_{k=0}^{n+1} {k\choose 1}{n+1-k\choose 1}={k+n+1-k+1\choose 1+1+1}=\frac{n(n+1)(n+2)}{6}$$
$$\begin{align} &1\cdot n + 2\cdot (n-1) + 3\cdot (n-3)+\cdots + (n-1)\cdot 2 + n\cdot1\\ &=\sum_{r=1}^n r(n+1-r)\\ &=\sum_{r=1}^n \sum_{j=1}^{n+1-r}r\\ &=\sum_{j=1}^n \sum_{r=1}^{n+1-j}r\\ &=\sum_{j=1}^n \sum_{r=1}^j {r\choose 1}\\ &=\sum_{j=1}^n {j+1\choose 2}\\ &={n+2 \choose 3}\qquad\blacksquare \end{align}$$
Use generating functions. Write as:
$\begin{align*} S_{n + 1} &= \sum_{0 \le k \le n + 1} k (n + 1 - k) \end{align*}$
You see this is a convolution, the coefficients of the sum:
$\begin{align*} S(z) &= \sum_{n \ge 0} S_n z^n \\ &= \left( \sum_{k \ge 0} k z^k \right)^2 \\ &= \frac{z^2}{(1 - z)^4} \end{align*}$
Using the generalized binomial theorem, you get the coefficients as:
$\begin{align*} S_{n + 1} &= [z^{n + 1}] S(z) \\ &= [z^{n - 1}] (1 - z)^{-4} \\ &= (-1)^{n - 1} \binom{-4}{n - 1} \\ &= \binom{n - 1 + 4 - 1}{4 - 1} \\ &= \frac{n (n + 1) (n + 2)}{6} \end{align*}$
