\[\sum_{r=1}^{9} \left( \frac{r+3}{2^{r}} \right).^{9}C_{r} = \alpha \left( \frac{3}{2} \right) ^{9} - \beta\] , $\alpha, \beta \in N$, then $(\alpha + \beta)^2$ is equal to
81
The problem asks us to evaluate the summation $S = \sum_{r=1}^{9} \left( \frac{r+3}{2^{r}} \right) \binom{9}{r}$ and express it in the form $\alpha \left( \frac{3}{2} \right)^{9} - \beta$, then find the value of $(\alpha + \beta)^2$.
Solving the Binomial Summation
We can split the summation into two parts:
$ S = \sum_{r=1}^{9} \frac{r}{2^r} \binom{9}{r} + \sum_{r=1}^{9} \frac{3}{2^r} \binom{9}{r} $Evaluating the Summation Parts
We use the binomial theorem: $\sum_{r=0}^{n} \binom{n}{r} x^r = (1+x)^n$. Let $n=9$ and $x=1/2$. Then $\sum_{r=0}^{9} \binom{9}{r} \left(\frac{1}{2}\right)^r = \left(1+\frac{1}{2}\right)^9 = \left(\frac{3}{2}\right)^9$.
- Part 1: Evaluate $\sum_{r=1}^{9} \frac{3}{2^r} \binom{9}{r}$. $ \sum_{r=1}^{9} \frac{3}{2^r} \binom{9}{r} = 3 \left( \sum_{r=0}^{9} \binom{9}{r} \left(\frac{1}{2}\right)^r - \binom{9}{0}\left(\frac{1}{2}\right)^0 \right) $ $ = 3 \left( \left(\frac{3}{2}\right)^9 - 1 \right) = 3 \left(\frac{3}{2}\right)^9 - 3 $
- Part 2: Evaluate $\sum_{r=1}^{9} \frac{r}{2^r} \binom{9}{r}$. We use the identity $r \binom{n}{r} = n \binom{n-1}{r-1}$. Here, $r \binom{9}{r} = 9 \binom{8}{r-1}$. $ \sum_{r=1}^{9} \frac{r}{2^r} \binom{9}{r} = \sum_{r=1}^{9} \frac{1}{2^r} \left( 9 \binom{8}{r-1} \right) = 9 \sum_{r=1}^{9} \frac{1}{2^r} \binom{8}{r-1} $ Let $k = r-1$. The sum becomes: $ 9 \sum_{k=0}^{8} \frac{1}{2^{k+1}} \binom{8}{k} = \frac{9}{2} \sum_{k=0}^{8} \binom{8}{k} \left(\frac{1}{2}\right)^k $ Using the binomial theorem with $n=8$ and $x=1/2$: $ \sum_{k=0}^{8} \binom{8}{k} \left(\frac{1}{2}\right)^k = \left(1+\frac{1}{2}\right)^8 = \left(\frac{3}{2}\right)^8 $ So, Part 2 is $\frac{9}{2} \left(\frac{3}{2}\right)^8$.
Combining and Simplifying
Now, combine the results for Part 1 and Part 2:
$ S = \frac{9}{2} \left(\frac{3}{2}\right)^8 + 3 \left(\frac{3}{2}\right)^9 - 3 $ $ S = \frac{9}{2} \frac{3^8}{2^8} + 3 \frac{3^9}{2^9} - 3 $ $ S = \frac{3^2 \cdot 3^8}{2^9} + \frac{3 \cdot 3^9}{2^9} - 3 $ $ S = \frac{3^{10}}{2^9} + \frac{3^{10}}{2^9} - 3 $ $ S = \frac{2 \cdot 3^{10}}{2^9} - 3 = \frac{3^{10}}{2^8} - 3 $Finding $\alpha$ and $\beta$
We need to write $S$ in the form $\alpha \left( \frac{3}{2} \right)^{9} - \beta$.
$ S = \frac{3^{10}}{2^8} - 3 = \frac{3 \cdot 3^9}{2^8} - 3 $ $ S = \frac{3 \cdot 3^9 \cdot 2}{2^9} - 3 = \frac{6 \cdot 3^9}{2^9} - 3 $ $ S = 6 \left(\frac{3}{2}\right)^9 - 3 $Comparing this with $\alpha \left( \frac{3}{2} \right)^{9} - \beta$, we find $\alpha = 6$ and $\beta = 3$. Both are natural numbers.
Calculating the Final Result
We need to find $(\alpha + \beta)^2$.
$ (\alpha + \beta)^2 = (6 + 3)^2 = 9^2 = 81 $Similar Questions
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