I am having difficulty to finish my proof.

Please help me finish and understand .

Suppose that $a \geq -1$ .Use induction to prove that $(1+a)^n\geq1+an $ for every $ n\in$ N.

Be sure to say where the assumption that $a\geq -1 $ is used.

Proof: For n= 1, $(1+a)^1\geq 1+1*a$ then it is true.

Suppose that the result n=k is holds.

Then we have $(1+a)^n\geq1+ka$ true.

Now I want to show that n=k+1 is true:

$(1+a)^{k+1}\geq 1+((k+1)*a)$ which is the same as :

$(1+a)^{k}(1+a)\geq 1+((k+1)*a)$

And here I am stuck.


If $(1+a)^k\ge1+ka$, then

$$(1+a)^k(1+a)\ge(1+ka)(1+a) \Rightarrow (1+a)^{k+1}\ge1+a+ka+ka^2$$

Observe that $ka^2\ge0$ ($a\ge-1$, so $a^2\ge0$, and $k\in\mathbb{N}$ so $k\ge0$).

  • $\begingroup$ How did you get $ka^2\geq 0$ $\endgroup$ – Hanna7777 Sep 6 '14 at 19:14
  • $\begingroup$ $a\ge-1$, so $a^2\ge0$, and $k\in\mathbb{N}$ so $k\ge0$. Therefore $ka^2\ge0$. $\endgroup$ – rae306 Sep 6 '14 at 19:25

$$(1+a)^{n+1} = (1 +a)(1+a)^n \ge (1 + a)(1 + na) = 1 + (n+1)a +na^2 \ge 1 + (n+1)a.$$


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