Generating Functions to solve $\sum_{k=0}^{m-1} {{k+2} \choose {k}}$

Question

I have to use generating functions to solve the sum

$$\sum_{k=0}^{m-1} {{k+2} \choose {k}}$$

Now I did this:

$\sum_{k=0}^{m-1} {{k+2} \choose {k}} = \sum_{k=0}^{m} {{k+2} \choose {k}} - {{m+2} \choose m} = \frac{1}{2} ([\sum_{k=0}^{m} {(k+1)(k+2)}] - (m+1)(m+2) )$ then I continued algebrically and got that it is $= \frac{(m+1)(m+2)}{6} = F(m)$

But what is worrying me, is that a solution with the help of Generating functions? I merely used Algebra to solve it, but in the end we get to a generating function.

Or is the way of solving it using generating functions much different?

Answer 1

In general if $a_0,a_1,\dots,a_n,\dots$ is a sequence and $s_n=\sum_{k=0}^n a_k$ then if:

$$f(z)=\sum_{n=0}^\infty a_nz^n$$ Then $$\frac{1}{1-z}f(z)=\sum_{n=0}^\infty s_nz^n$$

So, if $a_n=\binom{n+2}{n}=\binom{n+2}{2}$, what is $f(z)$?