`(.^(50)C_(1))^(2)+2(.^(50)C_(2))^(2)+3(.^(50)C_(3))^(2)+.....+50(.^(50)C_(50))^(2)=`

Options

A. `.^(100)C_(50)`
B. `50.(.^(100)C_(50))`
C. `25(.^(100)C_(50))`
D. `(2^(49))/(49!)(1.3.35…..99)`

Explanation

To solve the problem, we need to evaluate the expression:

\[ S = \sum_{k=1}^{50} k \cdot \binom{50}{k}^2 \]

### Step 1: Rewrite the Sum
We can rewrite the sum by using the identity \( k \cdot \binom{n}{k} = n \cdot \binom{n-1}{k-1} \). Therefore, we have:

\[ S = \sum_{k=1}^{50} k \cdot \binom{50}{k}^2 = \sum_{k=1}^{50} 50 \cdot \binom{49}{k-1} \cdot \binom{50}{k} \]

### Step 2: Change the Index of Summation
By changing the index of summation from \( k \) to \( j = k - 1 \), we get:

\[ S = 50 \cdot \sum_{j=0}^{49} \binom{49}{j} \cdot \binom{50}{j+1} \]

### Step 3: Use Vandermonde's Identity
We can use Vandermonde's identity, which states that:

\[ \sum_{j=0}^{r} \binom{m}{j} \cdot \binom{n}{r-j} = \binom{m+n}{r} \]

In our case, we can apply it with \( m = 49 \), \( n = 50 \), and \( r = 50 \):

\[ \sum_{j=0}^{49} \binom{49}{j} \cdot \binom{50}{j+1} = \binom{49 + 50}{50} = \binom{99}{50} \]

### Step 4: Substitute Back
Now substituting back into our expression for \( S \):

\[ S = 50 \cdot \binom{99}{50} \]

### Final Answer
Thus, the final result is:

\[ S = 50 \cdot \binom{99}{50} \]

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