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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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up vote 1 down vote accepted

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.

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how did you come up with multiplying by 1+p on both sides? o-o its indeed a good trick – John Lee Oct 19 '12 at 4:37
    
Thats what You want to proof. The $n+1$ case! – ILoveMath Oct 19 '12 at 4:38
    
can i prove this without multiplying n+1 on both sides? and do i use this trick for all Induction problems? – John Lee Oct 19 '12 at 4:42
    
    
@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 ?

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