If `sum_(r=1)^9 ((r+3)/2^r)(.^9C_r)=alpha(3/2)^9+beta`, then `alpha+beta` is equal to :

Options

A. 9
B. 3
C. 6
D. 2

Answer

To solve the problem, we need to evaluate the expression:

\[ \sum_{r=1}^{9} \frac{(r+3)}{2^r} \binom{9}{r} \]

and express it in the form \( \alpha \left( \frac{3}{2} \right)^9 + \beta \), then find \( \alpha + \beta \).

### Step 1: Break down the summation

We can separate the summation into two parts:

\[ \sum_{r=1}^{9} \frac{(r+3)}{2^r} \binom{9}{r} = \sum_{r=1}^{9} \frac{r}{2^r} \binom{9}{r} + \sum_{r=1}^{9} \frac{3}{2^r} \binom{9}{r} \]

### Step 2: Evaluate the second summation

The second summation can be simplified as follows:

\[ \sum_{r=1}^{9} \frac{3}{2^r} \binom{9}{r} = 3 \sum_{r=1}^{9} \binom{9}{r} \left( \frac{1}{2} \right)^r = 3 \left( 1 + \frac{1}{2} \right)^9 - 3 \]

Using the binomial theorem, we find:

\[ = 3 \left( \frac{3}{2} \right)^9 - 3 \]

### Step 3: Evaluate the first summation

For the first summation \( \sum_{r=1}^{9} \frac{r}{2^r} \binom{9}{r} \), we can use the identity \( r \binom{n}{r} = n \binom{n-1}{r-1} \):

\[ \sum_{r=1}^{9} \frac{r}{2^r} \binom{9}{r} = 9 \sum_{r=1}^{9} \frac{1}{2^r} \binom{8}{r-1} \]

Changing the index of summation (let \( k = r - 1 \)) gives:

\[ = 9 \sum_{k=0}^{8} \frac{1}{2^{k+1}} \binom{8}{k} = \frac{9}{2} \sum_{k=0}^{8} \binom{8}{k} \left( \frac{1}{2} \right)^k = \frac{9}{2} \left( 1 + \frac{1}{2} \right)^8 \]

Thus,

\[ = \frac{9}{2} \left( \frac{3}{2} \right)^8 \]

### Step 4: Combine both parts

Now we combine both parts:

\[ \sum_{r=1}^{9} \frac{(r+3)}{2^r} \binom{9}{r} = \frac{9}{2} \left( \frac{3}{2} \right)^8 + 3 \left( \frac{3}{2} \right)^9 - 3 \]

### Step 5: Factor out common terms

Factoring out \( \left( \frac{3}{2} \right)^8 \):

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

Calculating the expression inside the parentheses:

\[ \frac{9}{2} + \frac{9}{2} = 9 \]

Thus, we have:

\[ = 9 \left( \frac{3}{2} \right)^8 - 3 \]

### Step 6: Identify \( \alpha \) and \( \beta \)

Now we can write:

\[ = 9 \left( \frac{3}{2} \right)^9 + (-3) \]

This means \( \alpha = 9 \) and \( \beta = -3 \).

### Step 7: Calculate \( \alpha + \beta \)

Finally, we calculate:

\[ \alpha + \beta = 9 - 3 = 6 \]

Thus, the final answer is:

\[ \boxed{6} \]