Certainly! Let's tackle the problem you provided in part (b).

Problem Statement:

You are asked to show that: $\binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \cdots + \binom{n}{n-1} = 2^{n-1}$

Solution:

This expression involves the sum of binomial coefficients for odd indices in the binomial expansion. We can use the following approach:

1. Recognize the Binomial Expansion for $(1 + x)^n$: The binomial expansion of $(1 + x)^n$ is given by: $(1 + x)^n = \sum_{k=0}^n \binom{n}{k} x^k$

2. Substitute $x = 1$ and $x = -1$:

   • When $x = 1$, the expansion becomes: $(1 + 1)^n = \sum_{k=0}^n \binom{n}{k} = 2^n$ This sum includes all binomial coefficients from $k = 0$ to $k = n$.

   • When $x = -1$, the expansion becomes: $(1 - 1)^n = \sum_{k=0}^n \binom{n}{k} (-1)^k = 0$ Here, terms with even $k$ are positive, and terms with odd $k$ are negative.

3. Separate the Odd and Even Terms: Now, we can interpret these two results:

   • The sum of all coefficients (with $x = 1$) is $2^n$.
   • The alternating sum of coefficients (with $x = -1$) is $0$.

   This means: $\sum_{\text{even } k} \binom{n}{k} + \sum_{\text{odd } k} \binom{n}{k} = 2^n$ and $\sum_{\text{even } k} \binom{n}{k} - \sum_{\text{odd } k} \binom{n}{k} = 0$

4. Solve for the Sum of Odd Terms: Adding these two equations gives: $2 \sum_{\text{odd } k} \binom{n}{k} = 2^n$ Therefore, $\sum_{\text{odd } k} \binom{n}{k} = 2^{n-1}$

Conclusion:

This proves that: $\binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \cdots + \binom{n}{n-1} = 2^{n-1}$

Would you like further details on any part of this solution?

Here are five related questions to deepen your understanding:

1. How does the binomial theorem help in solving problems involving sums of binomial coefficients?
2. Why is the alternating sum of binomial coefficients equal to zero when $x = -1$?
3. How would you prove the sum of binomial coefficients for even indices?
4. Can this approach be extended to find other specific sums of binomial coefficients?
5. What are the implications of this result in probability or combinatorics?

Tip: Whenever dealing with sums of binomial coefficients, consider evaluating the binomial expansion at specific values for simplification.