Question

Using the principle of Mathematical Induction, prove the following for all $n$ and $N$
1) $3^n>2^n$

Answer 1

Let $P\left(n\right):3^n>2^n$

Step (i) for $n=1$

$P\left(1\right):3>2$

which is true for $n=1$

Step (ii)

Let it is true for $n=k$

So, $3^k>2^k$

Step (iii)

We have to prove that it is true for $n=k+1$ using step (ii)

Now $3^{k+1}=3^k\times 3$

$3^{k+1}>2^k\times 3$ using step (ii)

$3^{k+1}>2^k(2+1)$

$3^{k+1}>2^{k+1}+2^k$

Therefore, $3^{k+1}>2^{k+1}$

So it is true for $n=k+1$

therefore, it is true for all $n\in N$

Hence proved.