Friday, 10 August 2018

Proving with induction $(1-x)^n




Prove using induction that $\forall n\in\mathbb N, \forall x\in \mathbb R: 0



My attempt:




Base: for n=1:1x<11+x1x2<1, true since $0

Suppose the statement is true for n, prove for n+1:



(1x)n+1=(1x)(1x)ni.h<(1x)1+nx



Now I got stuck, maybe another induction to show that 1+nx+x<1+nx? Is there another way?



Moreover, I was told it's wrong to begin with (1x)n+1 and reach to 11+(n+1)x but why? Is it assuming what I need to prove?



Answer



Apply again the base case: 1x<11+x and that x2>0 to get



1x1+nx<1(1+nx)(1+x)=11+(n+1)x+nx2<11+(n+1)x.


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