Can anyone help me find the value of $$\sum_{s=1}^n \sum_{r=0}^s \binom{n}{s}\binom{s}{r}$$
So I edited this as some people were saying it was wrong, but now I think the question does make sense. Also the options given in the source of the question were $(a)3^n−1$ $(b)3^n+1$ $(c)3^n$ $(d)3(3^n−1)$
Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by $f(x)=\sum_{s=1}^{n}\sum_{r=0}^{s}\binom{n}{s}\binom{s}{r}x^{r}$. Observe that \begin{eqnarray*} & & f(x)\\ & = & \sum_{s=1}^{n}\binom{n}{s}\sum_{r=0}^{s}\binom{s}{r}x^{r}\\ & = & \sum_{s=1}^{n}\binom{n}{s}(1+x)^{s}\\ & = & -1+\sum_{s=0}^{n}\binom{n}{s}(1+x)^{s}\\ & = & -1+\left[1+(1+x)\right]^{n}\\ & = & -1+\left(2+x\right)^{n}. \end{eqnarray*} Putting $x=1$, we obtain $\sum_{s=1}^{n}\sum_{r=0}^{s}\binom{n}{s}\binom{s}{r}=-1+3^{n}$.
From Prove the identity $\sum_{k=0}^{n}\sum_{r=0}^{k} \binom{k}{r} \binom{n}{k} = 3^n$ it follows that the difference to your quest just is the different start index of the outer sum, then providing the single addend for $k=0$. - Thence calculating that, you get $$\sum_{k=0}^0\sum_{r=0}^k\left(k\atop r\right)\left(0\atop k\right) = \sum_{r=0}^0\left(0\atop r\right)\left(0\atop 0\right) =\left(0\atop 0\right)\left(0\atop 0\right) = 1\cdot 1 = 1$$
Thence your quest is just 1 off from the other one, resulting in your provided answer (a).
--- rk
