## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 11 - 11.4 - Mathematical Induction - 11.4 Exercises - Page 806: 13

#### Answer

The formula was proved for $n=1$ The formula is correct if n is changed by $n+1$

#### Work Step by Step

Let's prove the formula for $n=1$: $\frac{n}{2}(5n-1)=\frac{1}{2}(5(1)-1)=\frac{1}{2}(4)=2$ It is correct! Now, suppose that the formula is correct, that is: $2+7+12+17+...+(5n-3)=\frac{n}{2}(5n-1)$ Now, let's prove the formula for $n+1$: $2+7+12+17+...+(5n-3)+[5(n+1)-3]=$ $=[2+7+12+17+...+(5n-3)]+(5n+5-3)=\frac{n}{2}(5n-1)+(5n+2)=\frac{5n^2-n}{2}+\frac{2(5n+2)}{2}=\frac{5n^2-n+10n+4}{2}=\frac{5n^2+9n+4}{2}=\frac{5n^2+4n+5n+4}{2}=\frac{n(5n+4)+1(5n+4)}{2}=\frac{(n+1)(5n+4)}{2}=\frac{n+1}{2}[(5n+5)-1]=\frac{n+1}{2}[5(n+1)-1]$ That is exactly the given formula if $n$ is changed by $n+1$

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