I'm not sure what you're saying. Can you please clarify?

[John112]

I need a bit of help proving the following statement
(n + 2)^n ≤ (n + 1)^n+1 where n is a positive integer. The (n+2) and (n+1) bases are making it hard for me solve this. I tried several time, I can't get the inductive step. Can someone lend me a little hand here?

The base case is real simple with n=1; But I can't make the leap from n=k to to n=K+1.

 

[Andrax]

you proved it for 1
suppose that (n + 2)^n ≤ (n + 1)^n+1
and prove that ((n+1)+2)^n+1≤ ((n+1)+1)^(n+1)+1
i remember when i first learned about induction these cases kinda confused me as well just go back to the lesson and re read it if you don't get it

 

[Mark44]

I need a bit of help proving the following statement
(n + 2)^n ≤ (n + 1)^n+1 where n is a positive integer. The (n+2) and (n+1) bases are making it hard for me solve this. I tried several time, I can't get the inductive step. Can someone lend me a little hand here?

The base case is real simple with n=1; But I can't make the leap from n=k to to n=K+1.

Where's the template? PF rules require that you use the template when you post a problem.

you proved it for 1
suppose that (n + 2)^n ≤ (n + 1)^n+1
and prove that ((n+1)+2)^n+1≤ ((n+1)+1)^(n+1)+1
i remember when i first learned about induction these cases kinda confused me as well just go back to the lesson and re read it if you don't get it

John112 and Andrax - use parentheses!

(n + 1)^n+1 means (n + 1)ⁿ + 1

If you intend this as (n + 1)^{n + 1} then use LaTeX or the SUP tags or at the least, write it as (n + 1)^(n + 1).