if `(1+a)^(n)=.^(n)C_(0)+.^(n)C_(1)a++.^(n)C_(2)a^(2)+ . . .+.^(n)C_(n)a^(n)`, then prove that
`(.^(n)C_(1))/(.^(n)C_(0))+(2(.^(n)C_(2)))/(.^(n)C_(1))+(3(.^(n)C_(3)))/(.^(n)C_(2))+. . . +(n(.^(n)C_(n)))/(.^(n)C_(n-1))=` Sum of first n natural numbers.

Answer

To prove the given statement, we start with the binomial expansion of \((1 + a)^n\):

\[ (1 + a)^n = \binom{n}{0} + \binom{n}{1} a + \binom{n}{2} a^2 + \ldots + \binom{n}{n} a^n \]

We need to prove that:

\[ \frac{\binom{n}{1}}{\binom{n}{0}} + \frac{2 \cdot \binom{n}{2}}{\binom{n}{1}} + \frac{3 \cdot \binom{n}{3}}{\binom{n}{2}} + \ldots + \frac{n \cdot \binom{n}{n}}{\binom{n}{n-1}} = \text{Sum of first } n \text{ natural numbers} \]

### Step 1: Express each term using the property of binomial coefficients
Using the property of binomial coefficients, we know that:

\[ \frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n - (r - 1)}{r} = \frac{n - r + 1}{r} \]

### Step 2: Rewrite the left-hand side
We can rewrite the left-hand side of the equation as follows:

\[ \frac{\binom{n}{1}}{\binom{n}{0}} + \frac{2 \cdot \binom{n}{2}}{\binom{n}{1}} + \frac{3 \cdot \binom{n}{3}}{\binom{n}{2}} + \ldots + \frac{n \cdot \binom{n}{n}}{\binom{n}{n-1}} \]

Substituting the property of binomial coefficients:

\[ = 1 \cdot \frac{n}{1} + 2 \cdot \frac{n-1}{2} + 3 \cdot \frac{n-2}{3} + \ldots + n \cdot \frac{1}{n} \]

### Step 3: Simplify the expression
This simplifies to:

\[ = n + (n-1) + (n-2) + \ldots + 1 \]

### Step 4: Calculate the sum
The sum \(n + (n-1) + (n-2) + \ldots + 1\) can be calculated using the formula for the sum of the first \(n\) natural numbers:

\[ \text{Sum} = \frac{n(n + 1)}{2} \]

### Conclusion
Thus, we have shown that:

\[ \frac{\binom{n}{1}}{\binom{n}{0}} + \frac{2 \cdot \binom{n}{2}}{\binom{n}{1}} + \frac{3 \cdot \binom{n}{3}}{\binom{n}{2}} + \ldots + \frac{n \cdot \binom{n}{n}}{\binom{n}{n-1}} = \frac{n(n + 1)}{2} \]

This is indeed the sum of the first \(n\) natural numbers. Hence, the statement is proved.