Question

Use induction on n to prove that fir all n>=2, 2^n+3^n<5^n

Answer 1

Explanation:

We first prove the base case, which is proving that the inequality holds for n=2:

$2^2+3^2=4+9=13$

So $2^2+3^2 < 5^2$ and base case is proven.

We then do the inductive step, which is assuming that the inequality holds for n=k, and proving out of that that the inequality also holds for n=k+1:

Assume the inequality holds for n=k. This means that

$2^k+3^k$

Our goal is then to show that $2^{k+1}+3^{k+1}$.

We have that

$2^{k+1}+3^{k+1}=2\cdot 2^k+3\cdot 3^k$

(since we're assuming that $2^k+3^k$, we know that $3\cdot (2^k+3^k)$ ).

So $2^{k+1}+3^{k+1}$, and the inductive step is proven.

Therefore we can conclude by the principle of mathematical induction that for all $n \geq 2,~2^n+3^n$