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## Homework Statement

[tex]a> -1, (1+a)^n \geq 1+na[/tex]

## Homework Equations

[tex]a>-1[/tex]

[tex]a+1> 0[/tex]

## The Attempt at a Solution

If I let n=0, and then n=1 i get that 1≥1, and a+1≥ a+1.

Add to each side n+1

[tex](1+a)^{n+1}+(1+a)^{n} \geq 1+na+(1+a(n+1))[/tex]

Then...? Perhaps subtract the original expressions on each side and then evaluate if i can show that n+1-n is ≥ n+1-n

[tex](1+a)^{n+1}-(1+a)^n \geq 1+a(n+1)-(1+an)[/tex]

[tex]a(a+1)^n \geq a[/tex]

However, I don't really know what I'v done. for a>-1, is it obvious that a(a+1)^n≥a?

Or have I attacked the problem in a wrong way? I'm new to induction, and completely green when it comes to bigger than, geq, etc. instead of =