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Found in: Page 330

### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

# Prove that ${{\mathbf{1}}}^{{\mathbf{2}}}{\mathbf{-}}{{\mathbf{2}}}^{{\mathbf{2}}}{\mathbf{+}}{{\mathbf{3}}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbf{\cdots }}{\mathbf{+}}{\mathbf{\left(}}{\mathbf{-}}{\mathbf{1}}{\mathbf{\right)}}^{\mathbf{n}\mathbf{-}\mathbf{1}}{{\mathbit{n}}}^{{\mathbf{2}}}{\mathbf{=}}\frac{\mathbf{\left(}\mathbf{-}\mathbf{1}\mathbf{\right)}\mathbf{n}\mathbf{\left(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{\right)}}{\mathbf{2}}$ whenever n is a positive integer.

It is proved that ${1}^{2}-{2}^{2}+{3}^{2}-\cdots +\left(-1{\right)}^{n-1}{n}^{2}=\frac{\left(-1{\right)}^{n-1}n\left(n+1\right)}{2}$ .

See the step by step solution

## Step 1: Mathematical Induction

The principle of mathematical induction is to prove that P(n) is true for all positive integer n in two steps.

1. Basic step : To verify that P(1) is true.

2. Inductive step : To prove the conditional statement if P(k) is true then P(k+1) is true.

## Step 2: Proof by induction

Let P(n) be the statement ${1}^{2}-{2}^{2}+{3}^{2}-\cdots +\left(-1{\right)}^{n-1}{n}^{2}=\frac{\left(-1{\right)}^{n-1}n\left(n+1\right)}{2}$ . Let us prove by induction on n.

Base Case:

For n = 1 , the value of LHS is 1 and RHS is 1. Since, both are equal the statement is true for P(1) .

Induction Case:

Assume that the statement is true for P(k) then prove for P(k+1) .

Let P(k) be true then ${1}^{2}-{2}^{2}+{3}^{2}-\cdots +\left(-1{\right)}^{k-1}{k}^{2}=\frac{\left(-1{\right)}^{k-1}k\left(k+1\right)}{2}$ . Let us prove the statement is true for as follows:

$\begin{array}{r}{1}^{2}-{2}^{2}+{3}^{2}-\cdots +\left(-1{\right)}^{k-1}{k}^{2}+\left(-1{\right)}^{k}\left(k+1{\right)}^{2}=\frac{\left(-1{\right)}^{k-1}k\left(k+1\right)}{2}+\left(-1{\right)}^{k}\left(k+1{\right)}^{2}\\ =\frac{\left(-1{\right)}^{k-1}k\left(k+1\right)+2\left(-1{\right)}^{k}\left(k+1{\right)}^{2}}{2}\\ =\frac{\left(-1{\right)}^{k}\left(k+1\right)\left[\left(-1\right)k+2\left(k+1\right)\right]}{2}\\ =\frac{\left(-1{\right)}^{k}\left(k+1\right)\left[-k+2k+2\right]}{2}\end{array}$

Further, simplify the values as follows:

$\begin{array}{r}{1}^{2}-{2}^{2}+{3}^{2}-\cdots +\left(-1{\right)}^{k-1}{k}^{2}+\left(-1{\right)}^{k}\left(k+1{\right)}^{2}=\frac{\left(-1{\right)}^{k}\left(k+1\right)\left[-k+2k+2\right]}{2}\\ =\frac{\left(-1{\right)}^{k}\left(k+1\right)\left[k+2\right]}{2}\\ =\frac{\left(-1{\right)}^{\left(k+1\right)-1}\left(k+1\right)\left(\left(k+1\right)+1\right)}{2}\end{array}$

Thus, P(k+1) is also true.

Therefore, the statement P(n) is true for every positive integer n.