If `C_(r )=""^(n)C_(r )`, then `C_(1)+3C_(3)+5C_(5)+…=2C_(2) +4C_(4)+6C_(6)+…`

Answer

To solve the equation \( C_1 + 3C_3 + 5C_5 + \ldots = 2C_2 + 4C_4 + 6C_6 + \ldots \), we will use the properties of binomial coefficients and some algebraic manipulations.

### Step-by-Step Solution:

1. **Understand the Binomial Expansion**:
The binomial expansion of \( (1 + x)^n \) is given by:
\[ (1 + x)^n = C(n, 0) + C(n, 1)x + C(n, 2)x^2 + C(n, 3)x^3 + \ldots + C(n, n)x^n \]

2. **Differentiate the Binomial Expansion**:
Differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}[(1 + x)^n] = n(1 + x)^{n-1} \]
The right-hand side becomes:
\[ C(n, 1) + 2C(n, 2)x + 3C(n, 3)x^2 + \ldots + nC(n, n)x^{n-1} \]

3. **Evaluate at \( x = 1 \)**:
Substituting \( x = 1 \) into the differentiated equation gives:
\[ n(2^{n-1}) = C(n, 1) + 2C(n, 2) + 3C(n, 3) + \ldots + nC(n, n) \]

4. **Evaluate at \( x = -1 \)**:
Now, substitute \( x = -1 \):
\[ n(0) = C(n, 1) - 2C(n, 2) + 3C(n, 3) - 4C(n, 4) + \ldots + (-1)^n C(n, n) \]
This simplifies to:
\[ 0 = C(n, 1) - 2C(n, 2) + 3C(n, 3) - 4C(n, 4) + \ldots \]

5. **Combine the Results**:
Adding the results from steps 3 and 4, we can separate the terms:
- The sum of the coefficients with odd indices (1, 3, 5, ...) on the left side.
- The sum of the coefficients with even indices (2, 4, 6, ...) on the right side.

6. **Conclude the Equality**:
From the combined results, we can see that:
\[ C_1 + 3C_3 + 5C_5 + \ldots = 2C_2 + 4C_4 + 6C_6 + \ldots \]
This shows that the original statement is true.

### Final Statement:
Thus, the statement \( C_1 + 3C_3 + 5C_5 + \ldots = 2C_2 + 4C_4 + 6C_6 + \ldots \) is true.

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