## Precalculus (10th Edition)

Published by Pearson

# Chapter 12 - Sequences; Induction; the Binomial Theorem - 12.2 Arithmetic Sequences - 12.2 Assess Your Understanding - Page 814: 39

#### Answer

$n^2$

#### Work Step by Step

We have to determine the sum: $S=1+3+5+.....+(2n-1)$ We write each term: $1=2\cdot 1-1$ $3=2\cdot 2-1$ $5=2\cdot 3-1$ .............................. $2n-1=2\cdot n-1$ We notice that there are $n$ terms. We add the equations side by side: $1+3+5+...+(2n-1)=(2\cdot 1-1)+(2\cdot 2-1)+....+(2\cdot n-1)$ $S=(2\cdot 1+2\cdot 2+...+2\cdot n)-n\cdot 1$ $S=2(1+2+...+n)-n$ We use the formula: $1+2+...+k=\dfrac{k(k+1)}{2}$ $S=2\cdot\dfrac{n(n+1)}{2}-n$ $S=n(n+1)-n$ $S=n^2+n-n$ $S=n^2$

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