How to use Generating Function to find sum of series?

Question

How to use generating function to find the sum of series $1.2.3+2.3.4+3.4.5...+(r-1).r.(r+1)$

Answer 1

For this particular problem, we don't really need to use generating function. We can use the fact $$(r-1)r(r+1) = \frac14\left((r-1)r(r+1)(r+2) - (r-2)(r-1)r(r+1) \right)$$ and the sum is a telescoping one.

If one insists of using generating function, we can start from the OGF for constant sequence $1$: $$\sum_{k=0}^\infty z^k = \frac{1}{1-z}$$ For any positive integer $n$, multiply both sides by $z^n$ and differentiate $n$ times, we get $$ \sum_{k=0}^\infty \left(\prod_{i=1}^n (k+i)\right) z^k = \frac{d^n}{dz^n}\left[\frac{z^n}{1-z}\right] = \frac{d^n}{dz^n}\left[\frac{1}{1-z} - ( 1 + z + \cdots + z^{n-1} ) \right] = \frac{n!}{(1-z)^{n+1}} $$ Mulitply boths sides by $\frac{1}{1-z}$. Expand coefficients in LHS as a Cauchy product, we get

$$ \sum_{k=0}^\infty \left(\sum_{\ell=0}^k \prod_{i=1}^n(\ell+i)\right)z^k = \frac{n!}{(1-z)^{n+2}} = \frac{1}{n+1}\frac{d^{n+1}}{dz^{n+1}}\left[\frac{1}{1-z}\right] = \frac{1}{n+1}\sum_{k=0}^\infty \left(\prod_{i=1}^{n+1} (k+i)\right) z^k $$ Compare coefficients of $z^k$ on both sides, we get $$\sum_{\ell=0}^k \left(\prod_{i=1}^n (\ell+i)\right) = \frac{1}{n+1}\left(\prod_{i=1}^{n+1} (k+i)\right) $$ For $n = 3$ and $k = r-2$, this reduces to

$$1\cdot 2 \cdot 3 +2 \cdot 3 \cdot 4 +3 \cdot 4 \cdot 5 + \cdots + (r-1) r (r+1) = \frac14 (r-1)r(r+1)(r+2)$$