Statement-1 : ` sum_(r=0)^(n) r^(2) ""^(n)C_(r) x^(r) = n (n-1) x^(2) (1 + x)^(n-2) + nx (1 +x)^(n-1)`
Statement-2: `sum_(r=0)^(n) r^(2) ""^(n)C_(r) = n (n-1)2^(n-2)+ n2^(n-1)` .

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the given problem, we need to verify the two statements provided and show that they are true.

### Step 1: Analyze Statement 2
We start with Statement 2:
\[ \sum_{r=0}^{n} r^2 \binom{n}{r} = n(n-1)2^{n-2} + n2^{n-1} \]
We can express \( r^2 \) in terms of \( r \) and \( r(r-1) \):
\[ r^2 = r(r-1) + r \]
Thus, we can rewrite the summation:
\[ \sum_{r=0}^{n} r^2 \binom{n}{r} = \sum_{r=0}^{n} r(r-1) \binom{n}{r} + \sum_{r=0}^{n} r \binom{n}{r} \]

### Step 2: Simplify the First Summation
The first summation can be simplified using the identity:
\[ r(r-1) \binom{n}{r} = n(n-1) \binom{n-2}{r-2} \]
Thus:
\[ \sum_{r=0}^{n} r(r-1) \binom{n}{r} = n(n-1) \sum_{r=2}^{n} \binom{n-2}{r-2} = n(n-1) 2^{n-2} \]

### Step 3: Simplify the Second Summation
The second summation can be simplified using the identity:
\[ r \binom{n}{r} = n \binom{n-1}{r-1} \]
Thus:
\[ \sum_{r=0}^{n} r \binom{n}{r} = n \sum_{r=1}^{n} \binom{n-1}{r-1} = n 2^{n-1} \]

### Step 4: Combine the Results
Now we combine both results:
\[ \sum_{r=0}^{n} r^2 \binom{n}{r} = n(n-1)2^{n-2} + n2^{n-1} \]
This confirms that Statement 2 is true.

### Step 5: Analyze Statement 1
Now we check Statement 1:
\[ \sum_{r=0}^{n} r^2 \binom{n}{r} x^r = n(n-1)x^2(1+x)^{n-2} + nx(1+x)^{n-1} \]
Using the same approach as in Statement 2, we can rewrite \( r^2 \) and analyze the summation:
\[ \sum_{r=0}^{n} r^2 \binom{n}{r} x^r = \sum_{r=0}^{n} r(r-1) \binom{n}{r} x^r + \sum_{r=0}^{n} r \binom{n}{r} x^r \]

### Step 6: Apply the Identities
Using the identities we derived earlier:
1. For \( r(r-1) \):
\[ \sum_{r=0}^{n} r(r-1) \binom{n}{r} x^r = n(n-1)x^2(1+x)^{n-2} \]
2. For \( r \):
\[ \sum_{r=0}^{n} r \binom{n}{r} x^r = nx(1+x)^{n-1} \]

### Step 7: Combine the Results for Statement 1
Combining both results gives:
\[ \sum_{r=0}^{n} r^2 \binom{n}{r} x^r = n(n-1)x^2(1+x)^{n-2} + nx(1+x)^{n-1} \]
Thus, Statement 1 is also true.

### Conclusion
Both statements are true, and Statement 2 is not a correct explanation for Statement 1.