Question

In Problem, prove each statement.

${\left(1+a\right)}^n\ge 1+na$for$a>0$

Short answer

The given statement is shown.

Step-by-step solution

Step 1. Given information

${\left(1+a\right)}^n\ge 1+na$

Step 2. Prove the given statement.

Put $n=1$in $(1+a{)}^n\ge 1+na$to check if the statement is true for $n=1$.

width="121">(1+a)1≥1+(1)a1+a≥1+a

Thus, the statement is true for $n=1$

Assume that the statement is true for width="37">n=k

Thus, $(1+a{)}^k\ge 1+ka$.

Find whether $(1+a{)}^{k+1}\ge 1+(k+1)a$

$\begin{array}{rcl}(1+a{)}^{k+1}& =& (1+a{)}^k·(1+a)\\ & \ge & (1+ka)(1+a)\\ & \ge & 1+k{a}^2+a+ka\\ & \ge & (1+k)a+1+k{a}^2\\ & \ge & (1+k)a+1\end{array}$

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

Since both the conditions of the Principle of Mathematical Induction are satisfied, the statement