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As the title says Prove by mathematical induction that for n ≥ 1 it is true that

$(1 + a)^n \geq 1 + na$ for $a \geq 0$

Having no clue on how to solve this

Paolo Franchi
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Mikael Sierra
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  • Do you know what induction is? Have you done induction proofs before? – Bob Krueger Jan 13 '16 at 00:21
  • not much, just some introduction. Okay i will look into proof by induction of bernoullis inequality, Thank you. – Mikael Sierra Jan 13 '16 at 00:28
  • Without induction, you could notice that, if $g(x)=(1+x)^n-1-nx$, then $g(0)=0$ and $g'(x)=n(1+x)^{n-1}-n$ is non-negative for $x\ge 0$. –  Jan 13 '16 at 00:28

1 Answers1

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It is trivial in the $n=1$ case. Now assume it is true for some $n$, then $(1+a)^{n+1}=(1+a)^n(1+a) \geq (1+an)(1+a)=1+an+a+a^2n \geq 1+an+a=1+a(n+1)$ by our assumption that $a \geq 0$ and the inductive hypothesis. Hence we have our result. This is about as straight forward of an induction argument as you can have.

siegehalver
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