## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Section 10.4 - Mathematical Induction - Exercise Set - Page 1085: 12

#### Answer

See the explanation below.

#### Work Step by Step

We use mathematical induction as follows: Statement ${{S}_{1}}$ is: ${{S}_{1}}=3$ Then, simplifying on the right, obtain \begin{align} & \frac{n\left( n+5 \right)}{2}=\frac{\left( 1 \right)\left( 1+5 \right)}{2} \\ & =3 \end{align}. This statement shows that ${{S}_{1}}$ is true, Suppose ${{S}_{k}}$ is true. Using ${{S}_{k}},{{S}_{k+1}}$ from the expression, ${{S}_{k}}=3+4+5+....+\left( k+2 \right)=\frac{k\left( k+5 \right)}{2}$ Adding $\left( k+3 \right)$ on both sides as: \begin{align} & 3+4+5+....+\left( k+3 \right)=\frac{\left( k \right)\left( k+5 \right)}{2}+k+3 \\ & =\frac{{{k}^{2}}+7k+6}{2} \\ & =\frac{\left( k+1 \right)\left( k+6 \right)}{2} \\ & =\frac{\left( k+1 \right)\left( \left( k+1 \right)+5 \right)}{2} \end{align} Thus, ${{S}_{k+1}}$ is true. The result, ${{S}_{n}}=3+4+5+....+\left( n+2 \right)=\frac{n\left( n+5 \right)}{2}$ is true by mathematical induction. Hence, the provided statement is proved by mathematical induction.

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