`sum_(k=0)^10 (2^k+3)^(10)C_(k)=alpha*2^10+beta*3^10` then value of `alpha+beta`

Answer

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

\[ \sum_{k=0}^{10} (2^k + 3) \binom{10}{k} \]

We can break this sum into two separate parts:

\[ \sum_{k=0}^{10} (2^k + 3) \binom{10}{k} = \sum_{k=0}^{10} 2^k \binom{10}{k} + \sum_{k=0}^{10} 3 \binom{10}{k} \]

### Step 1: Evaluate the first sum

The first sum is:

\[ \sum_{k=0}^{10} 2^k \binom{10}{k} \]

This can be simplified using the Binomial Theorem, which states that:

\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]

Setting \(x = 2\) and \(n = 10\), we have:

\[ (1 + 2)^{10} = 3^{10} = \sum_{k=0}^{10} \binom{10}{k} 2^k \]

Thus,

\[ \sum_{k=0}^{10} 2^k \binom{10}{k} = 3^{10} \]

### Step 2: Evaluate the second sum

The second sum is:

\[ \sum_{k=0}^{10} 3 \binom{10}{k} \]

Since \(3\) is a constant, we can factor it out:

\[ 3 \sum_{k=0}^{10} \binom{10}{k} \]

Using the Binomial Theorem again, we know that:

\[ \sum_{k=0}^{10} \binom{10}{k} = (1 + 1)^{10} = 2^{10} \]

Thus,

\[ \sum_{k=0}^{10} 3 \binom{10}{k} = 3 \cdot 2^{10} \]

### Step 3: Combine the results

Now we can combine the results from both sums:

\[ \sum_{k=0}^{10} (2^k + 3) \binom{10}{k} = 3^{10} + 3 \cdot 2^{10} \]

### Step 4: Compare with the given expression

We are given that:

\[ \sum_{k=0}^{10} (2^k + 3) \binom{10}{k} = \alpha \cdot 2^{10} + \beta \cdot 3^{10} \]

From our previous result, we can identify:

\[ \alpha = 3 \quad \text{and} \quad \beta = 1 \]

### Step 5: Find \( \alpha + \beta \)

Now we can find:

\[ \alpha + \beta = 3 + 1 = 4 \]

Thus, the final answer is:

\[ \boxed{4} \]