Chapter 12, Problem 51RE

To determine

To show: The given statement is true for all natural numbers.

Answer to Problem 51RE

The statement is true for all natural numbers.

Explanation of Solution

Given: The given expression is $2+6+18+\mathrm{...}+2\cdot 3^{n-1}=3^n-1$ .

Substitute $1$ for $n$ .

  $\begin{array}{c}2\cdot 3^{1-1}=3^1-1\\ 2\cdot 1=3-1\\ 2=2\end{array}$

The statement is true for $n=1$

Assume that the statement is true for $n=k$ .

  $2+6+18+\mathrm{...}+2\cdot 3^{k-1}=3^k-1$

Check for $n=k+1$ .

  $\begin{array}{c}2+6+18+\mathrm{...}+2\cdot 3^{k-1}+2\cdot 3^{\left[\left(k+1\right)-1\right]}=3^k-1+2\cdot 3^k\\ =3^k\left(1+2\right)-1\\ =3^k\left(3\right)-1\\ =3^{k+1}-1\end{array}$

Thus, the statement is true for $k+1$ .

Both the conditions for mathematical induction are satisfied.

Thus, the statement is true for all natural numbers.