Use mathematical induction to prove that, for any integer n >= 1, n k=0 2k = 2n+1 1
Use mathematical induction to prove that, for any integer n >= 1, n k=0 2k = 2n+1 1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please help me solve the problem 2.
![**Proof 3:** First, we’ll do the base case. Suppose that \( n = 1 \). Then,
\[
\sum_{i=0}^{n} i = \sum_{i=1}^{n} i = 1 = \frac{1(1+1)}{2}.
\]
So, the statement is true for \( n = 1 \), completing the base case. Now we’ll do the inductive step. Suppose that the statement is true for \( n = k + 1 \). We’ll show that it’s also true for \( n = k \). We can calculate
\[
\sum_{i=0}^{k+1} i = \frac{(k+1)(k+2)}{2} \quad \text{(inductive hypothesis)}
\]
\[
= \frac{k(k+1)}{2} + \frac{2(k+1)}{2} \quad \text{(algebra)}.
\]
Since what’s left includes the formula for the case \( n = k \), this completes the inductive step.
**Problem 2**
Use mathematical induction to prove that, for any integer \( n \geq 1 \),
\[
\sum_{k=0}^{n} 2^k = 2^{n+1} - 1
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Transcribed Image Text:**Proof 3:** First, we’ll do the base case. Suppose that \( n = 1 \). Then,
\[
\sum_{i=0}^{n} i = \sum_{i=1}^{n} i = 1 = \frac{1(1+1)}{2}.
\]
So, the statement is true for \( n = 1 \), completing the base case. Now we’ll do the inductive step. Suppose that the statement is true for \( n = k + 1 \). We’ll show that it’s also true for \( n = k \). We can calculate
\[
\sum_{i=0}^{k+1} i = \frac{(k+1)(k+2)}{2} \quad \text{(inductive hypothesis)}
\]
\[
= \frac{k(k+1)}{2} + \frac{2(k+1)}{2} \quad \text{(algebra)}.
\]
Since what’s left includes the formula for the case \( n = k \), this completes the inductive step.
**Problem 2**
Use mathematical induction to prove that, for any integer \( n \geq 1 \),
\[
\sum_{k=0}^{n} 2^k = 2^{n+1} - 1
\]
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