If sum_{r=1}^9 left(frac{r+3}{2^r}right) cdot | vedclass

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(C) Given the expression $\sum_{r=1}^9 \left(\frac{r+3}{2^r}\right) \cdot {}^9C_r = \alpha \left(\frac{3}{2}\right)^9 - \beta$. We can split the summa

Vedclass · Accepted Answer

$81$

Vedclass · Answer

(C) Given the expression $\sum_{r=1}^9 \left(\frac{r+3}{2^r}\right) \cdot {}^9C_r = \alpha \left(\frac{3}{2}\right)^9 - \beta$. We can split the summation as: $\sum_{r=1}^9 \frac{r}{2^r} {}^9C_r + 3 \sum_{r=1}^9 \frac{1}{2^r} {}^9C_r$. Using the identity $r \cdot {}^nC_r = n \cdot {}^{n-1}C_{r-1}$,the first part becomes: $\sum_{r=1}^9 \frac{9 \cdot {}^8C_{r-1}}{2^r} = \frac{9}{2} \sum_{r=1}^9 {}^8C_{r-1} \left(\frac{1}{2}\right)^{r-1} = \frac{9}{2} \left(1 + \frac{1}{2}\right)^8 = \frac{9}{2} \left(\frac{3}{2}\right)^8 = 3 \left(\frac{3}{2}\right)^9$. The second part is: $3 \left[ \sum_{r=0}^9 {}^9C_r \left(\frac{1}{2}\right)^r - {}^9C_0 \left(\frac{1}{2}\right)^0 \right] = 3 \left[ \left(1 + \frac{1}{2}\right)^9 - 1 \right] = 3 \left(\frac{3}{2}\right)^9 - 3$. Adding both parts: $3 \left(\frac{3}{2}\right)^9 + 3 \left(\frac{3}{2}\right)^9 - 3 = 6 \left(\frac{3}{2}\right)^9 - 3$. Comparing with $\alpha \left(\frac{3}{2}\right)^9 - \beta$,we get $\alpha = 6$ and $\beta = 3$. Therefore,$(\alpha + \beta)^2 = (6 + 3)^2 = 9^2 = 81$.

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

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