Prove using induction that $\forall n\in\mathbb N, \forall x\in \mathbb R: 0
My attempt:
Base: for n=1:1−x<11+x⟺1−x2<1, true since $0
Suppose the statement is true for n, prove for n+1:
(1−x)n+1=(1−x)(1−x)ni.h<(1−x)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 (1−x)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: 1−x<11+x and that x2>0 to get
1−x1+nx<1(1+nx)(1+x)=11+(n+1)x+nx2<11+(n+1)x.
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