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

Answer

To evaluate the sum \( \sum_{r=1}^{n} \binom{n}{r} 2^r \), we can use the Binomial Theorem. The Binomial Theorem states that:

\[ (x + y)^n = \sum_{r=0}^{n} \binom{n}{r} x^r y^{n-r} \]

In our case, we can set \( x = 2 \) and \( y = 1 \). Thus, we have:

\[ (2 + 1)^n = \sum_{r=0}^{n} \binom{n}{r} 2^r 1^{n-r} \]

This simplifies to:

\[ 3^n = \sum_{r=0}^{n} \binom{n}{r} 2^r \]

Now, we can separate the sum into two parts:

\[ 3^n = \binom{n}{0} 2^0 + \sum_{r=1}^{n} \binom{n}{r} 2^r \]

Since \( \binom{n}{0} = 1 \) and \( 2^0 = 1 \), we have:

\[ 3^n = 1 + \sum_{r=1}^{n} \binom{n}{r} 2^r \]

To isolate the sum \( \sum_{r=1}^{n} \binom{n}{r} 2^r \), we subtract 1 from both sides:

\[ \sum_{r=1}^{n} \binom{n}{r} 2^r = 3^n - 1 \]

Thus, the final result is:

\[ \sum_{r=1}^{n} \binom{n}{r} 2^r = 3^n - 1 \]