Book edition 7th

Author(s) Kenneth H. Rosen

Pages 808 pages

ISBN 9780073383095

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Short Answer

Prove that $1^{2} - 2^{2} + 3^{2} - \dots + (- 1)^{n - 1} n^{2} = \frac{(- 1) n (n + 1)}{2}$ whenever n is a positive integer.

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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 $1^{2} - 2^{2} + 3^{2} - \dots + (- 1)^{n - 1} n^{2} = \frac{(- 1)^{n - 1} n (n + 1)}{2}$ . 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 $1^{2} - 2^{2} + 3^{2} - \dots + (- 1)^{k - 1} k^{2} = \frac{(- 1)^{k - 1} k (k + 1)}{2}$ . Let us prove the statement is true for as follows:

$\begin{aligned} 1^{2} - 2^{2} + 3^{2} - \dots + (- 1)^{k - 1} k^{2} + (- 1)^{k} (k + 1)^{2} = \frac{(- 1)^{k - 1} k (k + 1)}{2} + (- 1)^{k} (k + 1)^{2} \\ = \frac{(- 1)^{k - 1} k (k + 1) + 2 (- 1)^{k} (k + 1)^{2}}{2} \\ = \frac{(- 1)^{k} (k + 1) [(- 1) k + 2 (k + 1)]}{2} \\ = \frac{(- 1)^{k} (k + 1) [- k + 2 k + 2]}{2} \end{aligned}$

Further, simplify the values as follows:

$\begin{aligned} 1^{2} - 2^{2} + 3^{2} - \dots + (- 1)^{k - 1} k^{2} + (- 1)^{k} (k + 1)^{2} = \frac{(- 1)^{k} (k + 1) [- k + 2 k + 2]}{2} \\ = \frac{(- 1)^{k} (k + 1) [k + 2]}{2} \\ = \frac{(- 1)^{(k + 1) - 1} (k + 1) ((k + 1) + 1)}{2} \end{aligned}$

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

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

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Discrete Mathematics and its Applications

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Short Answer

Prove that $1^{2} - 2^{2} + 3^{2} - \dots + (- 1)^{n - 1} n^{2} = \frac{(- 1) n (n + 1)}{2}$ whenever n is a positive integer.

Step by Step Solution

Step 1: Mathematical Induction

Step 2: Proof by induction

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Prove that $1^{2} - 2^{2} + 3^{2} - \dots + (- 1)^{n - 1} n^{2} = \frac{(- 1) n (n + 1)}{2}$ whenever n is a positive integer.