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

Options

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

Explanation

To solve the problem, we need to verify the two statements provided:

### Statement 1:
\[ \sum_{r=0}^{n} r \cdot \binom{n}{r} = n \cdot 2^{n-1} \]

### Statement 2:
\[ \sum_{r=0}^{n} r \cdot \binom{n}{r} \cdot x^r = n \cdot (1 + x)^{n-1} \cdot x \]

### Step-by-Step Solution:

**Step 1: Verify Statement 1**

We start with the left-hand side of Statement 1:
\[ \sum_{r=0}^{n} r \cdot \binom{n}{r} \]

Using the identity \( r \cdot \binom{n}{r} = n \cdot \binom{n-1}{r-1} \), we can rewrite the sum:
\[ \sum_{r=0}^{n} r \cdot \binom{n}{r} = \sum_{r=1}^{n} n \cdot \binom{n-1}{r-1} \]

Notice that when \( r = 0 \), the term is zero, so we can start from \( r = 1 \).

**Step 2: Simplify the Sum**

We can factor out \( n \):
\[ = n \cdot \sum_{r=1}^{n} \binom{n-1}{r-1} \]

Changing the index of summation by letting \( k = r - 1 \) (thus \( r = k + 1 \)):
\[ = n \cdot \sum_{k=0}^{n-1} \binom{n-1}{k} \]

Using the binomial theorem, we know that:
\[ \sum_{k=0}^{n-1} \binom{n-1}{k} = 2^{n-1} \]

**Step 3: Final Result for Statement 1**

Substituting back, we have:
\[ \sum_{r=0}^{n} r \cdot \binom{n}{r} = n \cdot 2^{n-1} \]
Thus, Statement 1 is **true**.

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**Step 4: Verify Statement 2**

Now we check Statement 2:
\[ \sum_{r=0}^{n} r \cdot \binom{n}{r} \cdot x^r \]

Using the same identity for \( r \):
\[ = \sum_{r=1}^{n} n \cdot \binom{n-1}{r-1} \cdot x^r \]

Factoring out \( n \):
\[ = n \cdot \sum_{r=1}^{n} \binom{n-1}{r-1} \cdot x^r \]

Changing the index of summation again by letting \( k = r - 1 \):
\[ = n \cdot \sum_{k=0}^{n-1} \binom{n-1}{k} \cdot x^{k+1} \]

This can be rewritten as:
\[ = n \cdot x \cdot \sum_{k=0}^{n-1} \binom{n-1}{k} \cdot x^k \]

Using the binomial theorem:
\[ = n \cdot x \cdot (1 + x)^{n-1} \]

**Step 5: Final Result for Statement 2**

Thus, we conclude that:
\[ \sum_{r=0}^{n} r \cdot \binom{n}{r} \cdot x^r = n \cdot (1 + x)^{n-1} \cdot x \]
This confirms that Statement 2 is also **true**.

### Conclusion:
Both statements are true, and Statement 2 is a correct explanation of Statement 1.

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