Evaluate : `sum_(r = 1)^(n) ""^(n)C_(r) 2^r`

Answer

To evaluate the sum \( \sum_{r=1}^{n} \binom{n}{r} 2^r \), we can use the Binomial Theorem. Here’s a step-by-step solution:

### Step 1: Understand the Binomial Theorem
The Binomial Theorem states that:
\[ (x + y)^n = \sum_{r=0}^{n} \binom{n}{r} x^r y^{n-r} \]
For our case, we can set \( x = 2 \) and \( y = 1 \).

### Step 2: Apply the Binomial Theorem
Using the values of \( x \) and \( y \):
\[ (2 + 1)^n = \sum_{r=0}^{n} \binom{n}{r} 2^r 1^{n-r} \]
This simplifies to:
\[ 3^n = \sum_{r=0}^{n} \binom{n}{r} 2^r \]

### Step 3: Separate the Terms
The sum \( \sum_{r=0}^{n} \binom{n}{r} 2^r \) includes the term for \( r = 0 \) (which is \( \binom{n}{0} 2^0 = 1 \)). Therefore, we can express our original sum as:
\[ \sum_{r=1}^{n} \binom{n}{r} 2^r = \sum_{r=0}^{n} \binom{n}{r} 2^r - \binom{n}{0} 2^0 \]
This leads to:
\[ \sum_{r=1}^{n} \binom{n}{r} 2^r = 3^n - 1 \]

### Step 4: Final Result
Thus, the value of the sum \( \sum_{r=1}^{n} \binom{n}{r} 2^r \) is:
\[ \boxed{3^n - 1} \]