`3C_0+5C_1+7C_2+ +(2n+3)C_n=2^n(n+3)`

Options

A. `3^(n)(n+3)`
B. `2^(n)(n-3)`
C. `2^(n)(n+3)`
D. `3^(n)(n-3)`

Answer

To prove the equation \(3C_0 + 5C_1 + 7C_2 + \ldots + (2n+3)C_n = 2^n(n+3)\), we can follow these steps:

### Step 1: Understand the Binomial Coefficient
The binomial coefficient \(C_k\) (or \(\binom{n}{k}\)) represents the number of ways to choose \(k\) elements from a set of \(n\) elements. The expression can be rewritten using binomial coefficients:
\[ \sum_{k=0}^{n} (2k + 3) C_k = 3C_0 + 5C_1 + 7C_2 + \ldots + (2n+3)C_n \]

### Step 2: Break Down the Summation
We can separate the summation into two parts:
\[ \sum_{k=0}^{n} (2k + 3) C_k = \sum_{k=0}^{n} 2k C_k + \sum_{k=0}^{n} 3 C_k \]
The second summation simplifies as follows:
\[ \sum_{k=0}^{n} 3 C_k = 3 \sum_{k=0}^{n} C_k = 3 \cdot 2^n \]
since \(\sum_{k=0}^{n} C_k = 2^n\).

### Step 3: Evaluate the First Summation
For the first summation, we can use the identity \(k C_k = n C_{k-1}\):
\[ \sum_{k=0}^{n} 2k C_k = 2 \sum_{k=1}^{n} k C_k = 2 \sum_{k=1}^{n} n C_{k-1} = 2n \sum_{j=0}^{n-1} C_j = 2n \cdot 2^{n-1} \]
where we changed the index of summation from \(k\) to \(j\) (where \(j = k - 1\)).

### Step 4: Combine the Results
Now we can combine both parts:
\[ \sum_{k=0}^{n} (2k + 3) C_k = 2n \cdot 2^{n-1} + 3 \cdot 2^n \]
Factoring out \(2^n\):
\[ = 2^n \left(n + \frac{3}{2}\right) = 2^n(n + 3) \]

### Step 5: Conclusion
Thus, we have shown that:
\[ 3C_0 + 5C_1 + 7C_2 + \ldots + (2n+3)C_n = 2^n(n + 3) \]
This completes the proof.