Use induction to prove that Σmn1(2j + 1) = 3η2 3n²
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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![### Proof by Mathematical Induction
We are given the following statement to prove using induction:
\[
\sum_{j=n}^{2n-1} (2j + 1) = 3n^2
\]
#### Step 1: Base Case
First, we need to verify the base case where \( n = 1 \).
When \( n = 1 \):
\[
\sum_{j=1}^{2 \cdot 1 - 1} (2j + 1) = 2 \cdot 1 + 1 = 3
\]
And:
\[
3 \cdot 1^2 = 3
\]
Thus, the base case holds true:
\[
\sum_{j=1}^{1} (2j + 1) = 3 = 3 \cdot 1^2
\]
#### Step 2: Inductive Hypothesis
Assume that for some integer \( k \geq 1 \), the statement holds true:
\[
\sum_{j=k}^{2k-1} (2j + 1) = 3k^2
\]
#### Step 3: Inductive Step
Prove that if the statement holds for \( n = k \), it must also hold for \( n = k+1 \).
Consider:
\[
\sum_{j=k+1}^{2(k+1)-1} (2j + 1)
\]
Expressing the sum:
\[
\sum_{j=k+1}^{2(k+1)-1} (2j + 1) = \sum_{j=k+1}^{2k+1} (2j + 1)
\]
Break it into two parts:
\[
\left( \sum_{j=k}^{2k-1} (2j + 1) \right) + (2(2k) + 1) + (2(2k+1) + 1)
\]
Using the inductive hypothesis, replace the sum up to \( 2k-1 \):
\[
3k^2 + (4k + 1) + (4k + 3)
\]
Simplifying the sum:
\[
3k^2 + 4k + 1 + 4k + 3 = 3k^2 + 8k + 4
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3eb514bb-44ba-433b-9221-45dffb06f396%2F6d775694-87bb-4a51-bef4-d03165bfe45b%2Frt3g0gc_processed.png&w=3840&q=75)
Transcribed Image Text:### Proof by Mathematical Induction
We are given the following statement to prove using induction:
\[
\sum_{j=n}^{2n-1} (2j + 1) = 3n^2
\]
#### Step 1: Base Case
First, we need to verify the base case where \( n = 1 \).
When \( n = 1 \):
\[
\sum_{j=1}^{2 \cdot 1 - 1} (2j + 1) = 2 \cdot 1 + 1 = 3
\]
And:
\[
3 \cdot 1^2 = 3
\]
Thus, the base case holds true:
\[
\sum_{j=1}^{1} (2j + 1) = 3 = 3 \cdot 1^2
\]
#### Step 2: Inductive Hypothesis
Assume that for some integer \( k \geq 1 \), the statement holds true:
\[
\sum_{j=k}^{2k-1} (2j + 1) = 3k^2
\]
#### Step 3: Inductive Step
Prove that if the statement holds for \( n = k \), it must also hold for \( n = k+1 \).
Consider:
\[
\sum_{j=k+1}^{2(k+1)-1} (2j + 1)
\]
Expressing the sum:
\[
\sum_{j=k+1}^{2(k+1)-1} (2j + 1) = \sum_{j=k+1}^{2k+1} (2j + 1)
\]
Break it into two parts:
\[
\left( \sum_{j=k}^{2k-1} (2j + 1) \right) + (2(2k) + 1) + (2(2k+1) + 1)
\]
Using the inductive hypothesis, replace the sum up to \( 2k-1 \):
\[
3k^2 + (4k + 1) + (4k + 3)
\]
Simplifying the sum:
\[
3k^2 + 4k + 1 + 4k + 3 = 3k^2 + 8k + 4
\]
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