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Chapter 1: Problem 11

Conjecture a formula for the sum of the first \(n\) odd natural numbers \(1+3+\cdots+(2 n-1)\), and prove your formula by using Mathematical Induction.

Short Answer

Expert verified
The sum of the first \(n\) odd natural numbers is \(S_n = n^2\). This is proven by Mathematical Induction.

Step by step solution

01

Conjecture a formula

The sum of first \(n\) natural numbers can be calculated using the formula \(S_n = n^2\). Therefore, the formula for the sum of first \(n\) odd natural numbers is also \(S_n = n^2\).
02

Base Case

The base case is for \(n=1\). The formula gives \(S_1 = 1^2 = 1\), which corresponds to the sum of the first one odd number, as expected.
03

Inductive Step-Induction Hypothesis

Assume that the formula holds for some \(k\), i.e., \(S_k = k^2\).
04

Inductive step- Induction Conclusion

We need to show that the formula holds for \(k + 1\), that is \(S_{k+1} = (k+1)^2\). Considering the \(k+1\)st term, \(S_{k+1} = S_k + (2(k+1)-1)\), substituting the inductive hypothesis (i.e., \(S_k = k^2\)) we get, \(S_{k+1} = k^2 + 2k + 1\), which further simplifies to \(S_{k+1} = (k+1)^2\). This completes the Mathematical Induction proof.

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Most popular questions from this chapter

Find all natural numbers \(n\) such that \(n^{2}<2^{n}\). Prove your assertion.

The symmetric difference of two sets \(A\) and \(B\) is the set \(D\) of all elements that belong to either \(A\) or \(B\) but not both. Represent \(D\) with a diagram. (a) Show that \(D=(A \backslash B) \cup(B \backslash A)\). (b) Show that \(D\) is also given by \(D=(A \cup B) \backslash(A \cap B)\).

Let \(f(x):=x^{2}\) for \(x \in \mathbb{R}\), and let \(E:=\\{x \in \mathbb{R}:-1 \leq x \leq 0)\) and \(F:=(x \in \mathbb{R}: 0 \leq x \leq 1\\}\). Show that \(E \cap F=\\{0\\}\) and \(f(E \cap F)=\\{0\\}\), while \(f(E)=f(F)=\\{y \in \mathbb{R}: 0 \leq y \leq 1\\}\) Hence \(f(E \cap F)\) is a proper subset of \(f(E) \cap f(F)\). What happens if 0 is deleted from the sets \(E\) and \(F ?\)

Prove that \(n^{3}+5 n\) is divisible by 6 for all \(n \in \mathbb{N}\).

Prove that \(2 n-3 \leq 2^{n-2}\) for all \(n \geq 5, n \in \mathbb{N}\)
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