Use mathematical induction to prove.
a. 1 + 2 + 22 + 23 + … + 2𝑛 −1 = 2𝑛 − 1
b. 1 + 5 + 52 + ⋯ + 5𝑛−1 =
1
(5𝑛 − 1)
4

Let's tackle these problems one by one using mathematical induction.

### Problem a: Prove that
\[
1 + 2 + 2^2 + 2^3 + \dots + 2^{n-1} = 2^n - 1
\]

#### **Step 1: Base Case**
We start with \( n = 1 \).

For \( n = 1 \):
\[
\text{Left-hand side (LHS)} = 2^{1-1} = 2^0 = 1
\]
\[
\text{Right-hand side (RHS)} = 2^1 - 1 = 2 - 1 = 1
\]
So, the base case holds: \( 1 = 1 \).

#### **Step 2: Inductive Hypothesis**
Assume that the formula holds for some \( n = k \), i.e.,
\[
1 + 2 + 2^2 + \dots + 2^{k-1} = 2^k - 1
\]

#### **Step 3: Inductive Step**
We need to show that the formula also holds for \( n = k + 1 \). Consider:
\[
\text{LHS for } n = k+1: 1 + 2 + 2^2 + \dots + 2^{k-1} + 2^k
\]

Using the inductive hypothesis:
\[
= (2^k - 1) + 2^k
\]
\[
= 2^k - 1 + 2^k
\]
\[
= 2 \cdot 2^k - 1
\]
\[
= 2^{k+1} - 1
\]

This matches the right-hand side for \( n = k + 1 \). Therefore, by induction, the statement is true for all \( n \geq 1 \).

### Problem b: Prove that
\[
1 + 5 + 5^2 + \dots + 5^{n-1} = \frac{1}{4} \left(5^n - 1\right)
\]

#### **Step 1: Base Case**
We start with \( n = 1 \).

For \( n = 1 \):
\[
\text{LHS} = 5^{1-1} = 5^0 = 1
\]
\[
\text{RHS} = \frac{1}{4} \left(5^1 - 1\right) = \frac{1}{4} \times 4 = 1
\]
So, the base case holds: \( 1 = 1 \).

#### **Step 2: Inductive Hypothesis**
Assume the formula holds for \( n = k \), i.e.,
\[
1 + 5 + 5^2 + \dots + 5^{k-1} = \frac{1}{4} \left(5^k - 1\right)
\]

#### **Step 3: Inductive Step**
We need to show that the formula holds for \( n = k + 1 \). Consider:
\[
\text{LHS for } n = k+1: 1 + 5 + 5^2 + \dots + 5^{k-1} + 5^k
\]

Using the inductive hypothesis:
\[
= \frac{1}{4} \left(5^k - 1\right) + 5^k
\]
\[
= \frac{1}{4} \times 5^k - \frac{1}{4} + 5^k
\]
\[
= \frac{1}{4} \times 5^k + \frac{4}{4} \times 5^k - \frac{1}{4}
\]
\[
= \frac{1}{4} \times 5^k + \frac{4}{4} \times 5^k - \frac{1}{4} 
\]
\[
= \frac{5^k}{4} + \frac{4 \times 5^k}{4} - \frac{1}{4} 
\]
\[
= \frac{5^k(1 + 4) - 1}{4}
\]
\[
= \frac{5 \times 5^k - 1}{4} 
\]
\[
= \frac{5^{k+1} - 1}{4}
\]

This matches the right-hand side for \( n = k + 1 \). Therefore, by induction, the statement is true for all \( n \geq 1 \).

Would you like more details or have any questions?

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Here are 5 related questions:
1. How does the structure of a geometric series assist in solving these problems?
2. Can you generalize the method used here to prove other sums of powers?
3. What is the significance of the base case in mathematical induction?
4. How would the proof differ if the formula were incorrect?
5. How could you apply this technique to solve a sum involving negative powers?

**Tip:** When using mathematical induction, always ensure the base case is correct before proceeding to the inductive step, as the entire proof relies on this foundation.If you have any additional questions, want to explore similar topics, or need further clarification on these proofs, feel free to ask!

Use mathematical induction to prove: a) 1 + 2 + 2^2 + 2^3 + … + 2^n − 1 = 2^n − 1 b) 1 + 5 + 5^2 + ⋯ + 5^n−1 = (5^n − 1)/4

Explore how to prove the sums of geometric series using mathematical induction. This detailed solution covers the step-by-step process for proving the series 1 + 2 + 2^2 + ... + 2^n-1 and 1 + 5 + 5^2 + ... + 5^n-1. Suitable for students in Grades 11-12.

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