Find sum of `sum_(r=1)^n r . C (2n,r)` (a) `n*2^(2n-1)` (b) `2^(2n-1)` (c) `2^(n-1)+1` (d) None of these

Options

A. `n*2^(2n-1)`
B. `2^(2n-1)`
C. `2^(n-1)+1`
D. None of these

Answer

To find the sum \( \sum_{r=1}^{n} r \cdot \binom{2n}{r} \), we can use properties of binomial coefficients and some algebraic manipulation. Here’s a step-by-step solution:

### Step 1: Rewrite the Sum
We start with the expression:
\[ S = \sum_{r=1}^{n} r \cdot \binom{2n}{r} \]

### Step 2: Use the Identity for \( r \cdot \binom{n}{r} \)
We can use the identity:
\[ r \cdot \binom{n}{r} = n \cdot \binom{n-1}{r-1} \]
Applying this to our sum gives:
\[ S = \sum_{r=1}^{n} r \cdot \binom{2n}{r} = \sum_{r=1}^{n} 2n \cdot \binom{2n-1}{r-1} \]
This allows us to factor out \( 2n \):
\[ S = 2n \sum_{r=1}^{n} \binom{2n-1}{r-1} \]

### Step 3: Change the Index of Summation
Next, we change the index of summation. Let \( k = r - 1 \), then when \( r = 1 \), \( k = 0 \) and when \( r = n \), \( k = n - 1 \). Thus:
\[ S = 2n \sum_{k=0}^{n-1} \binom{2n-1}{k} \]

### Step 4: Use the Binomial Theorem
According to the binomial theorem, we know that:
\[ \sum_{k=0}^{m} \binom{m}{k} = 2^m \]
In our case, we have:
\[ \sum_{k=0}^{2n-1} \binom{2n-1}{k} = 2^{2n-1} \]
However, we only need the sum up to \( n-1 \). By symmetry of binomial coefficients:
\[ \sum_{k=0}^{n-1} \binom{2n-1}{k} = \frac{1}{2} \sum_{k=0}^{2n-1} \binom{2n-1}{k} = \frac{1}{2} \cdot 2^{2n-1} = 2^{2n-2} \]

### Step 5: Substitute Back into the Expression for \( S \)
Now substituting back into our expression for \( S \):
\[ S = 2n \cdot 2^{2n-2} = n \cdot 2^{2n-1} \]

### Final Answer
Thus, the sum \( S \) is:
\[ \boxed{n \cdot 2^{2n-1}} \]