`"if "(1+x)^(n)=C_(0)+C_(1).x+C_(2).x^(2)+C_(3).x^(3)+......+C_(n).x^(n),` then prove that
`C_(0)+2C_(1)+4C_(2)+6C_(3)+…….+2n.C_(n)=1+n.2^(n)`

Explanation

To prove that

\[ C_0 + 2C_1 + 4C_2 + 6C_3 + \ldots + 2nC_n = 1 + n \cdot 2^n, \]

we start with the binomial expansion of \((1 + x)^n\):

\[ (1 + x)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + \ldots + C_n x^n. \]

### Step 1: Define the sum we want to evaluate
Let

\[ S = C_0 + 2C_1 + 4C_2 + 6C_3 + \ldots + 2nC_n. \]

### Step 2: Rewrite the coefficients using symmetry
Using the property of binomial coefficients, we know that \(C_k = C_{n-k}\). Therefore, we can rewrite the sum \(S\) as follows:

\[ S = C_0 + 2C_1 + 4C_2 + 6C_3 + \ldots + 2nC_n = \sum_{k=0}^{n} 2k C_k. \]

### Step 3: Use a generating function
To evaluate \(S\), we can use the derivative of the binomial expansion. Consider:

\[ f(x) = (1 + x)^n. \]

Taking the derivative gives:

\[ f'(x) = n(1 + x)^{n-1}. \]

### Step 4: Evaluate at \(x = 1\)
Now, we can evaluate \(f'(1)\):

\[ f'(1) = n(1 + 1)^{n-1} = n \cdot 2^{n-1}. \]

### Step 5: Relate the derivative to the sum
The derivative \(f'(x)\) can also be expressed in terms of the coefficients:

\[ f'(x) = C_1 + 2C_2 x + 3C_3 x^2 + \ldots + nC_n x^{n-1}. \]

Evaluating at \(x = 1\) gives:

\[ f'(1) = C_1 + 2C_2 + 3C_3 + \ldots + nC_n. \]

### Step 6: Formulate the final expression
We can express \(S\) in terms of \(f'(1)\):

\[ S = 2(C_1 + 2C_2 + 3C_3 + \ldots + nC_n) = 2f'(1) = 2n \cdot 2^{n-1}. \]

### Step 7: Combine results
Now, we need to add \(C_0\) to our expression for \(S\):

\[ S = 2n \cdot 2^{n-1} + C_0. \]

Since \(C_0 = 1\) (the coefficient of \(x^0\) in the expansion of \((1 + x)^n\)), we have:

\[ S = 2n \cdot 2^{n-1} + 1. \]

### Step 8: Simplify the expression
Now, we can simplify:

\[ S = 1 + n \cdot 2^n. \]

### Conclusion
Thus, we have shown that

\[ C_0 + 2C_1 + 4C_2 + 6C_3 + \ldots + 2nC_n = 1 + n \cdot 2^n. \]