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Chapter 1: Problem 5

Prove that \(1^{2}-2^{2}+3^{2}+\cdots+(-1)^{n+1} n^{2}=(-1)^{n+1} n(n+1) / 2\) for all \(n \in \mathbb{N}\).

Short Answer

Expert verified
The relationship \(1^{2}-2^{2}+3^{2}+\cdots+(-1)^{n+1} n^{2}=(-1)^{n+1} n(n+1) / 2\) holds true for all \(n \in \mathbb{N}\), as proved by mathematical induction.

Step by step solution

01

Base Case (n=1)

At n=1, the left side of the equation becomes \( 1^{2} = 1\). The right side of the equation: \( (-1)^{1+1} \cdot 1 \cdot (1+1) / 2 = 1\). Then, both sides of the equation amounts to 1. Hence, the base case is true.
02

Induction Hypothesis

We assume the formula is true for some arbitrary natural number \(n\). That is, \(((-1)^{n+1} n^{2}=(-1)^{n+1} n(n+1) / 2\) is correct.
03

Induction Step

We need to prove that this formula is also true for \(n+1\). Take left side of the equation at \(n+1\): \(1^{2}-2^{2}+3^{2}+\cdots+(-1)^{n+2} (n+1)^{2}\). Split this into two series as: \((1^{2}-2^{2}+3^{2}+\cdots+(-1)^{n+1} n^{2}) + (-1)^{n+2} (n+1)^{2}\). We know from our assumption that the first part is equal to \( (-1)^{n+1} n(n+1) / 2 \). Substitute this into the equation to get: \( (-1)^{n+1} n(n+1) / 2 + (-1)^{n+2} (n+1)^{2} \). Simplify to get right side of the equation at \(n+1\): \( (-1)^{n+2}(n+1)(n+2) / 2 \). This completes the induction.

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