If _r=1^9r+3r2^-r^9C_r = (32)^9-, , N, then (+)^2 is equal to:

Question

If $\displaystyle\sum_{r=1}^{9}\binom{r+3}{r}\cdot2^{-r}\cdot{}^9C_r = \alpha\left(\frac32\right)^9-\beta$, $\alpha,\beta\in N$, then $(\alpha+\beta)^2$ is equal to:

Accepted Answer

Split $\binom{r+3}{2}\binom9r$-type sum using $r\cdot{}^9C_r=9\cdot{}^8C_{r-1}$:

$$\sum_{r=1}^{9}\binom{r+3}{2}\cdot{}^9C_r\left(\frac12\right)^r = \frac{9}{2}\sum_{r=1}^9{}^8C_{r-1}\left(\frac12\right)^{r-1}+3\sum_{r=1}^9{}^9C_r\left(\frac12\right)^r$$

$$=\frac92\left(1+\frac12\right)^8+3\left[\left(1+\frac12\right)^9-1\right]=\frac92\cdot\frac{3^8}{2^8}+3\left[\frac{3^9}{2^9}-1\right]$$

$$=\frac{3^{10}}{2^9}+\frac{3^{10}}{2^9}-3=4\cdot\frac{3^{10}}{2^{10}}-3=4\left(\frac32\right)^{10}-3=6\left(\frac32\right)^9-3$$

So $\alpha=6,\ \beta=3$, and $(\alpha+\beta)^2=81$.

Solutions · 1