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.
I think it's better to write $\sum_1^9$$\frac{r}{2^r}9C_r$+$\frac{3}{2^r}9C_r$ and now we have $$9C_r=\frac{9}{r}8C_{r-1}$$ and we also have $$(1+\frac{1}{2})^8=\sum_0^88C_r\frac{1}{2^r}$$ and $$(1+\frac{1}{2})^9=\sum_0^99C_r\frac{1}{2^r}$$ and from the question it's visible that $$3\sum_1^9\frac{1}{2^r}9C_r=3((1+\frac{1}{2})^9-1)$$ and i think you got the rest of the idea..now use this $$9C_r=\frac{9}{r}8C_{r-1}$$ and the first term translates to $$\sum_1^9\frac{r}{2^r}9C_r=\sum_{r=1}^9\frac{9}{2^r}8C_{r-1}$$ and $$\sum_{r=1}^9\frac{9}{2^r}8C_{r-1}=\frac{9}{2}(1+\frac{1}{2})^8$$ so the reqired answer is $$\frac{9}{2}(1+\frac{1}{2})^8+3((1+\frac{1}{2})^9-1)$$ and $\alpha=6$ $\beta=-3$ which gives $\alpha$+$\beta$=3 Hope that Helps!
With the help of Thenard, I am able to solve this question. Here is my answer: (it's basically Thenard's answer. I am just writing it in a different way so as to make it easier for future reference.)
$$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 $3$.