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I've just started going through some very basic analysis and was working through an attempt at a proof that $(1+a)^n \geq 1 + na + \frac{1}{2} n (n-1)a^2 $

I got about half way through, and found I was having some trouble figuring out the last few steps. So looked at the solution. Solution

It all seems to make sense apart from the third line where the $+ \frac{1}{2}n(n-1)a^3$ term seems to have disapppeared as '$a^3 > 0$' (which I do no understand.)

I carried out the initial step, by subbing in n=1, and then during my induction steps I had already managed to get the last line via replacing $n+1$ in the original expression, so was trying to equate with the expansion of $(1+a)^n (1+a)$ and though seemed to have gotten lost trying to get there.

Could anyone please explain what has happened at the third line there?

Thanks

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    $\begingroup$ If $a>0$ then $x+a^3>x$ ... $\endgroup$ Commented Oct 1, 2015 at 12:00
  • $\begingroup$ Ah... That simple. Thanks... $\endgroup$ Commented Oct 1, 2015 at 12:18

1 Answer 1

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Third step goes like this –

$$(1+a)^{n+1}\geq(1+a)(1+na)+ \tfrac{1}{2} n(n-1) a^2 + \tfrac12 n(n-1) a^3\\ \geq (1+a)(1+na)+\tfrac12n(n-1) a^2\ \ (\text{as } a^3>0)$$

Hope I got what you are trying to ask..

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