- #1
bhoom
- 15
- 0
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 =