<s>[INST] Evaluate $\sum\limits_{1 \le j\le k\le n} jk $ I am trying to solve an exercise from Concrete Mathematics but I seem to be stuck on the sum $$\sum_{1 \le j\le k\le n} jk $$ How to proceed? I have tried using Iverson's bracket condition like 
$$ [1\le j\le k\le n] = [1\le j \le n][j\le k \le n] $$
but I am not sure how to write the sum as multiple sums.
 [/INST] An alternative is to set $j-1=a, k-j=b, n-k=c$. Then we are looking for
$$ S(n)=\sum_{\substack{a+b+c=n-1\\a,b,c\geq 0}}(a+1)(a+b+1)=\sum_{\substack{a+b+c=n-1\\a,b,c\geq 0}}(1+2a+b+a^2+ab) $$
where
$$ \sum_{d\geq 0}x^d = \frac{1}{1-x},\qquad \sum_{d\geq 0}dx^d = \frac{x}{(1-x)^2},\qquad \sum_{d\geq 0}d^2 x^d = \frac{x(1+x)}{(1-x)^3} $$
and
$$ \sum_{\substack{a+b+c=n-1\\a,b,c\geq 0}}\!\!\!1 = [x^{n-1}]\frac{1}{(1-x)^3},\qquad \sum_{\substack{a+b+c=n-1\\a,b,c\geq 0}}a=[x^{n-1}]\frac{x}{(1-x)^4} $$
$$ \sum_{\substack{a+b+c=n-1\\a,b,c\geq 0}}\!\!\!a^2 = [x^{n-1}]\frac{x(1+x)}{(1-x)^5},\qquad \sum_{\substack{a+b+c=n-1\\a,b,c\geq 0}}ab=[x^{n-1}]\frac{x^2}{(1-x)^5} $$
so
$$ S(n) = [x^{n-1}]\left(\frac{1}{(1-x)^3}+\frac{3x}{(1-x)^4}+\frac{x(1+x)}{(1-x)^5}+\frac{x^2}{(1-x)^5}\right)$$
or
$$ S(n) = [x^{n-1}]\frac{2x+1}{(1-x)^5}=\binom{n+3}{4}+2\binom{n+2}{4}=\color{red}{\frac{n(n+1)(n+2)(3n+1)}{24}}. $$

An interesting and more practical way is to prove first that $S(n)$ is a fourth-degree polynomial in the $n$ variable, then find its coefficients through Lagrange interpolation, once computed $S(1),S(2),S(3),S(4),S(5)$.
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