TITLE: $\sum_{r=1}^9(\frac{r+3}{2^r}){9\choose r}=\alpha(\frac32)^9+\beta$
QUESTION [0 upvotes]: If $\sum_{r=1}^9(\frac{r+3}{2^r}){9\choose r}=\alpha(\frac32)^9+\beta$ then $\alpha+\beta=?$

Opening the sum, I get $\frac42{9\choose1}+\frac5{2^2}{9\choose2}+...+\frac{12}{2^{9}}{9\choose9}$.
I see it's a combination of A.P. and G.P. but not able to proceed next.

REPLY [0 votes]: With the help of Thenard, I am able to solve this question. Here is my answer: (it's basically Thenard's answer. I am just writing it in a different way so as to make it easier for future reference.) 
$$LHS=\sum_{r=1}^{9}\frac r{2^r}{9\choose r}+3\sum_{r=1}^{9}\frac 1{2^r}{9\choose r}$$
$$=\frac92\sum_{r=1}^{9}\frac 1{2^{r-1}}{8\choose r-1}+3[(1+\frac12)^9-1]$$
$$=\frac92(1+\frac12)^8+3(\frac32)^9-3$$
$$=(3+3)(\frac32)^9-3$$
Comparing with RHS, we get $\alpha=6,\beta=-3$. So, the answer is $3$.
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