# how: mathematical induction prove inequation

Provided that $p\geq-1$, prove $(1+p)^n\geq1+np$ for all integers $n\geq 0$

Also, where in the calculation do I use $p\geq -1$?

Thanks guy!

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Please, add homework tag here. –  Nikita Evseev Oct 19 '12 at 4:29

For base case, if $n = 0$, then $(1 + p)^0 = 1 \geq 1$. So Equation trivially holds! No suppose it holds for $n$ in $Z^+$. Then $$(1+p)^{n+1} = (1+p)^n(1+p) \geq (1+np)(1+p) = 1+np^2 + np + p \geq 1 +(n+1)p.$$ The proof is concluded by math induction.
Thats what You want to proof. The $n+1$ case! –  ILoveMath Oct 19 '12 at 4:38
@LJym89 The term $n^2p$ should be $np^2$ (note that $n^2p$ could be negative). –  Erick Wong Oct 19 '12 at 4:49
I think you use this : $(1+p)^n \geq 1 + np$ then $(1+p)^n(1+p) \geq (1+np)(1+p)$ if $p < -1$ then the inequality reversed, that is your question, right ?