Understanding the Problem: Binomial Expansion

We are asked to find the sum of the coefficients of the terms with odd powers of x in the expansion of \( (1 + x)^{50} \). This involves understanding the binomial theorem and properties of binomial coefficients.

Let's first recall the binomial theorem for the expansion of \( (1 + x)^n \):

\( (1 + x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{k}x^k + \dots + \binom{n}{n}x^n \)

In this formula, \( \binom{n}{k} \) represents the binomial coefficient for the term \( x^k \).

Expansion of (1 + x)⁵⁰

For the given problem, the exponent is \( n = 50 \). The expansion of \( (1 + x)^{50} \) is:

\( (1 + x)^{50} = \binom{50}{0} + \binom{50}{1}x + \binom{50}{2}x^2 + \binom{50}{3}x^3 + \dots + \binom{50}{49}x^{49} + \binom{50}{50}x^{50} \)

Identifying Coefficients of Odd and Even Powers

The coefficients of the odd powers of x are the coefficients of \( x^1, x^3, x^5, \dots, x^{49} \). These are \( \binom{50}{1}, \binom{50}{3}, \binom{50}{5}, \dots, \binom{50}{49} \).

Let \( S_{odd} \) be the sum of these coefficients:

\( S_{odd} = \binom{50}{1} + \binom{50}{3} + \binom{50}{5} + \dots + \binom{50}{49} \)

The coefficients of the even powers of x are the coefficients of \( x^0, x^2, x^4, \dots, x^{50} \). These are \( \binom{50}{0}, \binom{50}{2}, \binom{50}{4}, \dots, \binom{50}{50} \).

Let \( S_{even} \) be the sum of these coefficients:

\( S_{even} = \binom{50}{0} + \binom{50}{2} + \binom{50}{4} + \dots + \binom{50}{50} \)

Using Properties of Binomial Coefficients

A standard technique to find sums involving binomial coefficients is to substitute specific values for x in the binomial expansion.

Sum of All Coefficients (Setting x = 1)

If we set \( x = 1 \) in the expansion of \( (1 + x)^{50} \), we get the sum of all coefficients:

\( (1 + 1)^{50} = \binom{50}{0} + \binom{50}{1}(1) + \binom{50}{2}(1)^2 + \dots + \binom{50}{50}(1)^{50} \)

\( 2^{50} = \binom{50}{0} + \binom{50}{1} + \binom{50}{2} + \dots + \binom{50}{50} \)

So, the sum of all coefficients is \( S_{even} + S_{odd} = 2^{50} \). Let's call this Equation (1).

Alternating Sum of Coefficients (Setting x = -1)

If we set \( x = -1 \) in the expansion of \( (1 + x)^{50} \), the terms with odd powers of x will become negative, and terms with even powers will remain positive:

\( (1 + (-1))^{50} = \binom{50}{0} + \binom{50}{1}(-1) + \binom{50}{2}(-1)^2 + \binom{50}{3}(-1)^3 + \dots + \binom{50}{50}(-1)^{50} \)

\( (0)^{50} = \binom{50}{0} - \binom{50}{1} + \binom{50}{2} - \binom{50}{3} + \dots + \binom{50}{50} \)

Since \( 50 > 0 \), \( (0)^{50} = 0 \). Rearranging the terms:

\( 0 = (\binom{50}{0} + \binom{50}{2} + \dots + \binom{50}{50}) - (\binom{50}{1} + \binom{50}{3} + \dots + \binom{50}{49}) \)

\( 0 = S_{even} - S_{odd} \)

This gives us \( S_{even} = S_{odd} \). Let's call this Equation (2).

Solving for the Sum of Odd Power Coefficients

We have a system of two linear equations with two variables, \( S_{even} \) and \( S_{odd} \):

1. \( S_{even} + S_{odd} = 2^{50} \)
2. \( S_{even} - S_{odd} = 0 \)

From Equation (2), we know that \( S_{even} = S_{odd} \). Substitute this into Equation (1):

\( S_{odd} + S_{odd} = 2^{50} \)

\( 2 \cdot S_{odd} = 2^{50} \)

Now, we can solve for \( S_{odd} \) by dividing both sides by 2:

\( S_{odd} = \frac{2^{50}}{2} \)

Using the rule of exponents \( \frac{a^m}{a^n} = a^{m-n} \):

\( S_{odd} = 2^{50 - 1} \)

\( S_{odd} = 2^{49} \)

Conclusion: Sum of Coefficients of Odd Powers

The sum of the coefficients of odd powers of x in the expansion of \( (1 + x)^{50} \) is \( 2^{49} \).

Revision Table: Key Binomial Properties

Additional Information: Generalization

For any binomial expansion of \( (1+x)^n \), where \( n \) is a positive integer, the sum of the coefficients of the even powers of x is equal to the sum of the coefficients of the odd powers of x. Both sums are equal to \( 2^{n-1} \).

This can be derived using the same method as above:

• \( (1+1)^n = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \binom{n}{3} + \dots = 2^n \)
• \( (1-1)^n = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + \dots = 0 \) (for \( n > 0 \))

Adding these two equations gives:

\( 2 \left( \binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \dots \right) = 2^n \)

Sum of even power coefficients \( = \frac{2^n}{2} = 2^{n-1} \).

Subtracting the second equation from the first gives:

\( 2 \left( \binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \dots \right) = 2^n \)

Sum of odd power coefficients \( = \frac{2^n}{2} = 2^{n-1} \).

Thus, for \( (1+x)^{50} \), where \( n=50 \), the sum of odd power coefficients is \( 2^{50-1} = 2^{49} \).