Chapter 12, Problem 52RE

To determine

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

Answer to Problem 52RE

The statement is true for all natural numbers.

Explanation of Solution

Given: The given expression is $3+6+12+\mathrm{...}+3\cdot 2^{n-1}=3\left(2^n-1\right)$ .

Substitute $1$ for $n$ .

  $\begin{array}{c}3\cdot 2^{1-1}=3\left(2^1-1\right)\\ 3\cdot 1=3\cdot \left(2-1\right)\\ 3=3\end{array}$

The statement is true for $n=1$

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

  $3+6+12+\mathrm{...}+3\cdot 2^{k-1}=3\left(2^k-1\right)$

Check for $n=k+1$ .

  $\begin{array}{c}3+6+12+\mathrm{...}+3\cdot 2^{\left(k+1\right)-1}=3\left(2^k-1\right)+3\cdot 2^k\\ =\left(3\cdot 2^k+3\cdot 2^k\right)-3\\ =2\left(3.\cdot 2^k\right)-3\\ =2\left(2^{k+1}-1\right)\end{array}$

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.