Question

Prove by the Principle of Mathematical Induction: $n^2<2^n$for all natural numbers $n\ge 5$.

Answer 1

Assume given statement

Let $P\left(n\right):n^2<2^n\forall n\ge 5,nϵN$

Check that statement is true for $n=5$

$P\left(1\right):5^2<2^5$

$=25<32$

$\Rightarrow P\left(n\right)$ is true for $n=5$

Assume P(k) to be true and then prove $P(k+1)$ is true.

Lets assume P(k) is true.

$k^2<2^k\dots \left(1\right)$

To prove: $(k+1{)}^2<2^{(k+1)}$

LHS$=(k+1{)}^2$

$=k^2+2k+1<2^k+2k+1$ $\{$from (1)$\}$

Now we know that $2k+1<2^k$for all $k>5$ as LHS will be increased by 2 for every term but RHS is twice of the last term

$\therefore k^2+2k+1<2^k+2^k$

$\Rightarrow (k+1{)}^2<{2.2}^k$

$\Rightarrow (k+1{)}^2<2^{k+1}$

Hence, $P(k+1)$ is true whenever P(k) is true.

Hence, By Principle of mathematical Induction P(n) is true for all natural numbers $n\ge 5.$