# Thread: Proof by induction question 2

1. ## Proof by induction question 2

Hi guys

Need help with this proof buy induction.

Let $x\in REAL$ with $x\geq-1$. Prove that $(1+x)^n\geq1+nx$ for every integer $n\geq1$
SOLUTION:

Let $P(n)$ be the statement that $(1+x)^n\geq1+nx$

anchor step: $P(1)$
obviously true

induction step:

suppose $x\geq1$ is an integer and that $P(n)$ is true
since $x\geq-1$ we have $1+x\geq0$ and so using $P(n)$ gives

$(1+x)^{n+1}=(1+x)\cdot(1+x)^n$
$\geq (1+x)(1+nx)$
$=1+x+nx+nx^2=1+(n+1)x+nx^2 \geq 1+(n+1)x$

so we have deduced $P(n+1)$ . Hence $P(n)$ holds for all $n \geq 1$ .

ok here's the problem.
i understand $(1+x)^{n+1}=(1+x)\cdot(1+x)^n$ by law of exponentials
but then this : $\geq (1+x)(1+nx)$ ???? I mean where does this come from: $(1+x)(1+nx)$
obviously in the hypothesis we have $(1+x)^n\geq1+nx$ ... But how to get from $(1+x)^n\geq1+nx$ to $(1+x)^{n+1} \geq (1+x)(1+nx)$ ?? I tried to sub-in $n+1$ and it didnt work..

$(1+(n+1)x) = nx + x + 1 = ????$

How to get from here to $(1+x)(1+nx)$ ????

Obviously im missing something

Thank u guys

2. ## Re: Proof by induction question 2

use the binomial expansion

$(1+x)^n = \displaystyle{\sum_{k=0}^n}~\dfrac{x^k}{k!}$

3. ## Re: Proof by induction question 2

Originally Posted by shakra
Need help with this proof buy induction.
If $x>-1$ then $(1+x)^n\ge (1+nx)$
SOLUTION:
obviously true $n=1$

induction step: $(1+x)^n\ge(1+nx)$
\begin{align*}(1+x)^{1+n}&=(1+x)(1+x)^{n}\\&\ge(1+ x)(1+nx)\\&\ge 1+nx+x+nx^2\\&\ge 1+nx+x\\&\ge 1+(n+1)x \end{align*}

4. ## Re: Proof by induction question 2

Originally Posted by shakra
Hi guys

Need help with this proof buy induction.

SOLUTION:

obviously true

induction step:

ok here's the problem.
i understand $(1+x)^{n+1}=(1+x)\cdot(1+x)^n$ by law of exponentials
but then this : $\geq (1+x)(1+nx)$ ???? I mean where does this come from: $(1+x)(1+nx)$You are assuming that P(n) is true ie $(1+x)^n >= 1 + nx$ so that $(1+x)*(1+x)^n >=(1+x)(1+nx)$
obviously in the hypothesis we have $(1+x)^n\geq1+nx$ ... But how to get from $(1+x)^n\geq1+nx$ to $(1+x)^{n+1} \geq (1+x)(1+nx)$ ?? I tried to sub-in $n+1$ and it didnt work..

$(1+(n+1)x) = nx + x + 1 = ????$

How to get from here to $(1+x)(1+nx)$ ????

Obviously im missing something

Thank u guys
see comment in red