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### 1. Problem 1:
For \( n \geq 1 \), prove the following identity:
\[
\binom{n}{0} + 2\binom{n}{1} + 2^2\binom{n}{2} + \cdots + 2^n\binom{n}{n} = 3^n
\]

#### Solution:
We can solve this using the **Binomial Theorem**.

The Binomial Theorem states:
\[
(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k
\]

Now, if we choose \( x = 1 \) and \( y = 2 \), then:
\[
(1 + 2)^n = \sum_{k=0}^{n} \binom{n}{k} 1^{n-k} 2^k
\]
Since \( 1^{n-k} = 1 \) for all \( k \), we simplify this to:
\[
3^n = \sum_{k=0}^{n} \binom{n}{k} 2^k
\]
This matches the left-hand side of the expression in the problem:
\[
\binom{n}{0} + 2\binom{n}{1} + 2^2\binom{n}{2} + \cdots + 2^n\binom{n}{n}
\]
Thus, the identity is proven:
\[
\binom{n}{0} + 2\binom{n}{1} + 2^2\binom{n}{2} + \cdots + 2^n\binom{n}{n} = 3^n
\]

### 2. Problem 2:
For \( n \geq 1 \), prove:
\[
\binom{n}{r} < \binom{n}{r+1}
\]
if and only if \( 0 \leq r < \frac{1}{2}(n-1) \).

#### Solution:
We want to prove that for \( 0 \leq r < \frac{n-1}{2} \), the binomial coefficient satisfies \( \binom{n}{r} < \binom{n}{r+1} \).

Using the property of binomial coefficients, we know that:
\[
\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n-r}{r+1}
\]

Now, we analyze this ratio:
- If \( \frac{n-r}{r+1} > 1 \), then \( \binom{n}{r+1} > \binom{n}{r} \).
- Simplifying \( \frac{n-r}{r+1} > 1 \) gives us:
\[
n - r > r + 1
\]
\[
n > 2r + 1
\]
\[
r < \frac{n-1}{2}
\]
Thus, the condition \( r < \frac{n-1}{2} \) ensures that \( \binom{n}{r+1} > \binom{n}{r} \).

Therefore, we have proven that \( \binom{n}{r} < \binom{n}{r+1} \) if and only if \( 0 \leq r < \frac{1}{2}(n-1) \).

---

Do you have any questions or would you like further details on any of these solutions?

### Follow-up Questions:
1. Can you explain how the Binomial Theorem is used in general?
2. How would you apply the Binomial Theorem to different values of \( x \) and \( y \)?
3. What other methods could be used to prove inequality in binomial coefficients?
4. Can this inequality between binomial coefficients hold for other ranges of \( r \)?
5. How does the Binomial Coefficient change with different values of \( n \) and \( r \)?

### Tip:
Remember that binomial coefficients have a symmetric property: \( \binom{n}{r} = \binom{n}{n-r} \), which can simplify many binomial expressions!

Untuk n ≥ 1, buktikan (n0) + 2(n1) + 2^2(n2) + ... + 2^n(n n) = 3^n. Buktikan untuk n ≥ 1 bahwa (n r) < (n r+1) jika dan hanya jika 0 ≤ r < 1/2(n-1).

Prove that for n ≥ 1, (n0) + 2(n1) + 2^2(n2) + ... + 2^n(n n) = 3^n, and also prove for n ≥ 1 that (n r) < (n r+1) if and only if 0 ≤ r < 1/2(n-1). Detailed steps using Binomial Theorem and algebraic properties of coefficients.

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