Hi, I have the following problem which I think I have solved correctly (it would be great if you could check it), but I don't know how to answer to "simple" questions about my proof... any help would be very much appreciated, thanks!

The problem says:

"Let t>-1 be given with t≠0. Show by induction that for n≥2 we have

$\displaystyle (1+t)^t>1+nt$."

And here are the two questions that I don't know:

"Where is the condition t>-1 used in your proof? Is the result true if instead we had -2<t<-1?"

My solution to the first bit is:

Let $\displaystyle F(n)=(1+t)^n>1+nt$.

Proof by induction on n.

Base n=0. Then $\displaystyle (1+t)^n=(1+t)^0=1=1+0t=1+nt$

Induction. Suppose that there is a number k (k≥2) such thaqt $\displaystyle (1+t)^k>1+kt$ for any t>-1 (t≠0). We need to show that (1+t)^{k+1}>1+(k+1)t for any t>-1, t≠0.

$\displaystyle (1+t)^{k+1}=(1+t)^k*(1+t)$. By the induction hypothesis, $\displaystyle (1+t)^k>1+kt$, so we have:

$\displaystyle (1+t)^k*(1+t) \geq (1+kt)*(1+t)=1+kt+t+kt^2=1+(1+k)t+kt^2$.

Since $\displaystyle t^2>0$ and $\displaystyle k \geq 2, 1+(1+k)t+kt^2 \geq 1+(1+k)t$.

So $\displaystyle (1+t)^{k+1} \geq 1+(k+1)t$

And so the result holds for n=0 and hence for all n≥2 and t>-1,t≠0.