## Precalculus (6th Edition) Blitzer

Let us assume ${{S}_{n}}$ be a statement involving the positive integer n. If 1. ${{S}_{1}}$ is true, and 2. Let us assume that the ${{S}_{k}}$ be true for some integer $k$. Then, one has to use ${{S}_{k}}$ to prove ${{S}_{k+1}}$ is also true. If one is able to prove ${{S}_{k+1}}$ then by the principle of mathematical induction the statement ${{S}_{n}}$ is true for all positive integers $n$. Firstly, show that ${{S}_{1}}$ is true. For ${{S}_{1}}$ one has And write ${{S}_{1}}$ by taking its term on left and substituting n with $1$ on the right. ${{S}_{1}}:\sum\limits_{i=1}^{1}{5\cdot {{6}^{i}}=6\left( {{6}^{1}}-1 \right)}$ So, $5\cdot 6=5\cdot 6$ Therefore, the statement ${{S}_{1}}$ is true. Then, assume that ${{S}_{k}}$ is true, then ${{S}_{k+1}}$ should also be true by the method of mathematical induction. ${{S}_{k}}:\sum\limits_{i=1}^{k}{5\cdot {{6}^{i}}=6\left( {{6}^{k}}-1 \right)}$ Now, one has to prove ${{S}_{k+1}}$ is true ${{S}_{k+1}}:\sum\limits_{i=1}^{k+1}{5\cdot {{6}^{i}}=6\left( {{6}^{k+1}}-1 \right)}$ Since, one assume that ${{S}_{k}}$ is true, $\sum\limits_{i=1}^{k}{5\cdot {{6}^{i}}=6\left( {{6}^{k}}-1 \right)}$ By adding by $5\cdot {{6}^{k+1}}$ on both sides, \begin{align} & \sum\limits_{i=1}^{k}{5\cdot {{6}^{i}}+\left( 5\cdot {{6}^{k+1}} \right)=6\left( {{6}^{k}}-1 \right)}+\left( 5\cdot {{6}^{k+1}} \right) \\ & \sum\limits_{i=1}^{k+1}{5\cdot {{6}^{i}}=6\left( {{6}^{k}}-1 \right)}+\left( \left( 6-1 \right)\cdot {{6}^{k+1}} \right) \\ & \sum\limits_{i=1}^{k+1}{5\cdot {{6}^{i}}=\left( {{6}^{k+1}}-6 \right)}+\left( {{6}^{k+2}}-{{6}^{k+1}} \right) \\ & \sum\limits_{i=1}^{k+1}{5\cdot {{6}^{i}}=}\left( {{6}^{k+2}}-6 \right) \end{align} So, $\sum\limits_{i=1}^{k+1}{5\cdot {{6}^{i}}=6\left( {{6}^{k+1}}-1 \right)}$ So, the final statement is ${{S}_{k+1}}$. Thus, by the method of principal of mathematical induction, the statement $\sum\limits_{i=1}^{n}{5\cdot {{6}^{i}}=6\left( {{6}^{n}}-1 \right)}$ is true for every positive integer $n$.