`underset(r=1)overset(n)(sum)r(.^(n)C_(r)-.^(n)C_(r-1))` is equal to

Options

A. `2^(n)+1n+1`
B. `2^(n)-n+1`
C. `n-2^(n)+1`
D. `n-2^(n)-1`

Answer

To solve the problem \( \sum_{r=1}^{n} r \left( \binom{n}{r} - \binom{n}{r-1} \right) \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the given summation:
\[ \sum_{r=1}^{n} r \left( \binom{n}{r} - \binom{n}{r-1} \right) \]
This can be rewritten as:
\[ \sum_{r=1}^{n} r \binom{n}{r} - \sum_{r=1}^{n} r \binom{n}{r-1} \]

### Step 2: Simplify the first term
Using the identity \( r \binom{n}{r} = n \binom{n-1}{r-1} \), we can simplify the first term:
\[ \sum_{r=1}^{n} r \binom{n}{r} = n \sum_{r=1}^{n} \binom{n-1}{r-1} \]
The summation \( \sum_{r=1}^{n} \binom{n-1}{r-1} \) is equal to \( 2^{n-1} \) (the sum of the binomial coefficients for \( n-1 \)):
\[ \sum_{r=1}^{n} r \binom{n}{r} = n \cdot 2^{n-1} \]

### Step 3: Simplify the second term
For the second term, we can also apply the same identity:
\[ \sum_{r=1}^{n} r \binom{n}{r-1} = \sum_{r=0}^{n-1} (r+1) \binom{n}{r} = \sum_{r=0}^{n-1} r \binom{n}{r} + \sum_{r=0}^{n-1} \binom{n}{r} \]
The first part simplifies to:
\[ \sum_{r=0}^{n-1} r \binom{n}{r} = n \cdot 2^{n-2} \]
And the second part is simply \( 2^{n} \) (the sum of the binomial coefficients for \( n \)):
\[ \sum_{r=0}^{n-1} \binom{n}{r} = 2^{n} \]

### Step 4: Combine the results
Now, we can combine both results:
\[ \sum_{r=1}^{n} r \left( \binom{n}{r} - \binom{n}{r-1} \right) = n \cdot 2^{n-1} - \left(n \cdot 2^{n-2} + 2^{n}\right) \]
This simplifies to:
\[ = n \cdot 2^{n-1} - n \cdot 2^{n-2} - 2^{n} \]
\[ = n \cdot 2^{n-2} - 2^{n} \]
\[ = 2^{n-2} (n - 2) \]

### Step 5: Final result
Thus, the final result of the summation is:
\[ \sum_{r=1}^{n} r \left( \binom{n}{r} - \binom{n}{r-1} \right) = n - 2^{n+1} \]