Question

Need some help with mathematical induction

Answer 1

\[(1+p)^n \ge 1+np~~~for~~~p \ge-1\]

Answer 2

Checking n= 1 we get \[(1+p)^1 \ge 1+1* p\] which is a true statement so if I let n = n+1 then I have \[(1+p)^{(n+1)} \ge 1+np +(n+1)?\] but this doesn't make sense to me, maybe if I added (p+1) actually I'm not sure at all.

Answer 3

I just need a hint I think

Answer 4

@ganeshie8

Answer 5

If you're trying to prove by induction that \((1+p)^n\ge1+np\), then your proof should be using the assumption that this holds for \(n\) to show that \((1+p)^{n+1}\ge1+(n+1)p\).

Given that \((1+p)^n\ge1+np\), you know that
\[\begin{align*}
(1+p)^{n+1}&=(1+p)(1+p)^n\\[1ex]
&\ge(1+p)(1+np)&\color{lightgray}{\text{why?}}\\[1ex]
&=1+p+np+np^2\\[1ex]
&=1+(n+1)p+np^2\\[1ex]
&~~\vdots
\end{align*}\]

Answer 6

Accidentally noticed a proof that is NOT induction, might be nice to see anyways so here ya go. It uses the binomial theorem.

\[(1+p)^n = \sum_{k=0}^n \binom{n}{k} p^k \ge \sum_{k=0}^1 \binom{n}{k} p^k = 1+np\]

Answer 7

You know it's funny at first I started like @holsteremission but I thought we weren't allowed to do that for some reason, haha, and why, I'm not sure I'll get back to you once I figure it out! Thank you!

Answer 8

Just multiply both sides by (1+p) eh