Let's solve the two parts of this problem.

Part (i): Prove $\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} = 2^n$

This is a well-known result from the Binomial Theorem.

Binomial Theorem:

The binomial expansion of $(1 + x)^n$ is: $(1 + x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n$

Substitute $x = 1$:

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

Simplify: $2^n = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n}$

Thus, the sum of all binomial coefficients is $2^n$, proving part (i).

Part (ii): Prove $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \dots + (-1)^n \binom{n}{n} = 0$

This result also follows from the Binomial Theorem.

Binomial Theorem:

The binomial expansion of $(1 - x)^n$ is: $(1 - x)^n = \binom{n}{0} - \binom{n}{1}x + \binom{n}{2}x^2 - \dots + (-1)^n \binom{n}{n}x^n$

Substitute $x = 1$:

$(1 - 1)^n = \binom{n}{0} - \binom{n}{1}(1) + \binom{n}{2}(1^2) - \dots + (-1)^n \binom{n}{n}(1^n)$

Simplify: $0^n = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \dots + (-1)^n \binom{n}{n}$

Since $0^n = 0$ for $n > 0$, the alternating sum of the binomial coefficients is $0$, proving part (ii).

Summary:

1. Part (i): The sum of all binomial coefficients equals $2^n$.
2. Part (ii): The alternating sum of binomial coefficients equals $0$.

Would you like detailed step-by-step derivations or a visual explanation of the Binomial Theorem? Let me know!

Related Questions:

1. How does the Binomial Theorem relate to combinatorics?
2. Can we generalize these results for multinomials?
3. How does the proof change if $n = 0$ or $n = 1$?
4. What are real-world applications of the Binomial Theorem in probability?
5. Can we derive these results graphically using Pascal’s Triangle?

Tip:

The Binomial Theorem is essential for understanding combinatorics and expansions, so practice substituting various values of $x$ to observe how binomial coefficients behave.