Explanation:
We first prove the base case, which is proving that the inequality holds for n=2:
[tex]2^2+3^2=4+9=13<25=5^2 [/tex]
So [tex] 2^2+3^2 < 5^2[/tex] 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
[tex]2^k+3^k<5^k[/tex]
Our goal is then to show that [tex]2^{k+1}+3^{k+1}<5^{k+1}[/tex].
We have that
[tex]2^{k+1}+3^{k+1}=2\cdot 2^k+3\cdot 3^k<3\cdot 2^k+3\cdot 3^k =3\cdot (2^k+3^k)<3\cdot 5^k < 5\cdot 5^k = 5^{k+1}[/tex]
(since we're assuming that [tex]2^k+3^k<5^k[/tex], we know that [tex]3\cdot (2^k+3^k)<3\cdot 5^k[/tex] ).
So [tex]2^{k+1}+3^{k+1}<5^{k+1}[/tex], and the inductive step is proven.
Therefore we can conclude by the principle of mathematical induction that for all [tex] n \geq 2,~2^n+3^n<5^n [/tex]
