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Chapter 11: Problem 1

Write out in full and verify the statements \(S_{1}, S_{2}, S_{1} S_{4}\) and \(S_{5}\) for the following. Then use mathematical induction to prove that each statement is inue for every positive integer \(n .\) See Example 1 $$1+3+5+\cdots+(2 n-1)=n^{2}$$

Short Answer

Expert verified
The statement is true for all positive integers n, verified by induction.

Step by step solution

01

- Verify Statement S鈧�

Plug in the smallest positive integer value, which is 1: For n = 1, the statement becomes: \(1 = 1^2\)Since both sides are equal, this verifies that S鈧� is true.
02

- Verify Statement S鈧�

Plug in the value n = 2 into the statement: For n = 2, the sum is: \(1 + 3 = 4\) and \(2^2 = 4\)Again, since both sides are equal, S鈧� is true.
03

- Verify Statement S鈧�

Plug in the value n = 4 into the statement: For n = 4, the sum is: \(1 + 3 + 5 + 7 = 16\) and \((4)^2 = 16\)Since both sides are equal, S鈧� is true.
04

- Verify Statement S鈧�

Plug in the value n = 5 into the statement: For n = 5, the sum is: \(1 + 3 + 5 + 7 + 9 = 25\) and \((5)^2 = 25\)Since both sides are equal, S鈧� is true.
05

- Basis Step for Induction

We have already shown that S鈧� is true for n = 1 in Step 1.
06

- Inductive Step

Assume that the statement is true for some positive integer k, i.e., \(1 + 3 + 5 + \cdots + (2k - 1) = k^2\)This is the inductive hypothesis.
07

- Prove for k + 1

We need to show that the statement holds for k + 1: \(1 + 3 + 5 + \cdots + (2k - 1) + [2(k + 1) - 1] = (k + 1)^2\)Using the inductive hypothesis, this becomes: \(k^2 + [2(k + 1) - 1] = (k + 1)^2\)Simplify the right-hand side and left-hand side: \(k^2 + 2k + 1 = (k + 1)^2\)Since both sides are equal, the statement holds for k + 1.
08

- Conclusion

Since the statement is true for n = 1 (basis step) and has been shown to be true from n = k to n = k + 1 (inductive step), by mathematical induction, the statement is true for all positive integers n.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

sequence summation
In mathematics, 'sequence summation' refers to the process of adding up all the terms in a sequence. In our example, the sequence is made up of odd numbers: 1, 3, 5, and so on, up to \(2n-1\). The goal is to show that the sum of these numbers equals \(n^{2}\), where 'n' is the number of terms in the sequence.
Let's break it down further:
  • The first term (=1) is simply 1, which equals \(1^{2}\).
  • For =2, the sum of the first two odd numbers (1+3) is 4, which equals \(2^{2}\).
  • For =3, the sum of the first three odd numbers (1+3+5) is 9, which equals \(3^{2}\).
These examples show that the pattern \(1 + 3 + 5 + \cdots + (2n-1) = n^{2}\) holds for specific small values of 'n'. To prove it holds for all positive integers, we use a technique called mathematical induction.
inductive hypothesis
An 'inductive hypothesis' is the assumption that a statement is true for some arbitrary positive integer 'k'. This forms the base for proving that the statement is true for the next integer, \k+1\.
In our current example, the inductive hypothesis assumes that the equation:
\(1 + 3 + 5 + \cdots + (2k-1) = k^{2}\)
is true. This assumption allows us to proceed to the next step of the proof, which is showing it holds for \k+1\.
verification steps
Verification steps are used to confirm the initial cases and to show that our statement holds for the first few values. Here, we verify for \ = 1, 2, 4,\ and \5\:
  • **For \=1\:** The sequence sum is 1. \(n^{2}=1^{2}=1\).
  • **For \=2\:** The sequence sum is 1+3=4. \(n^{2}=2^{2}=4\).
  • **For \=4\:** The sequence sum is 1+3+5+7=16. \(n^{2}=4^{2}=16\).
  • **For \=5\:** The sequence sum is 1+3+5+7+9=25. \(n^{2}=5^{2}=25\).
With these verifications, we move on to the basis step and inductive step in mathematical induction. This allows us to verify our statement for all positive integers systematically.

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