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I've been able to follow the idea and steps of induction so far but I've hit a road block in understanding one of the examples in a text book. This is what the book says p.97:

Prove: $3^n \geq1 + 2n$

Skipping past the base case and assuming it's true, the books inductive step is as follows:

Show: $3^{n+1} = 1 + 2(1+n)$

LHS $= 3\cdot 3^n $
LHS $\geq 3(1+2n)$ [by assumption]
LHS $\geq 1+2+2n+4n$ [algebra]
LHS $\geq 1+2(1+n)$ [since n>0]

How can the $4n$ be omitted by $n>0$? This really boggles me, appreciate any insights and help.

Thanks!

2 Answers 2

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If n>0 then 4n>0 and omitting it from the RHS reduces the value. So >= still applies - or applies more strongly, and the = could be dropped.

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    Apologies for ambiguity there - I was not referring to the base case in dropping the equality, but to the inductive step. It is exploring such things which leads to extensions and generalisations.2012-03-03
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Since $n>0$, we have $4n>0$ so $1+2+2n+4n>1+2+2n=1+2(1+n)$, thus $\text{LHS}\geq 1+2(1+n)$.