Results 1 to 4 of 4

Thread: Proof by induction question 2

  1. #1
    Newbie
    Joined
    Jun 2016
    From
    london
    Posts
    16

    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
    Last edited by shakra; Nov 10th 2016 at 06:55 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    5,565
    Thanks
    2348

    Re: Proof by induction question 2

    use the binomial expansion

    $(1+x)^n = \displaystyle{\sum_{k=0}^n}~\dfrac{x^k}{k!}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,016
    Thanks
    2553
    Awards
    1

    Re: Proof by induction question 2

    Quote Originally Posted by shakra View Post
    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*}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Oct 2009
    From
    Brisbane
    Posts
    890
    Thanks
    199

    Re: Proof by induction question 2

    Quote Originally Posted by shakra View Post
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof by induction question
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: Nov 11th 2016, 12:06 PM
  2. simple proof by induction question
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: Jan 9th 2011, 09:44 PM
  3. proof by induction question involving power set
    Posted in the Discrete Math Forum
    Replies: 16
    Last Post: Jan 22nd 2010, 08:18 AM
  4. Proof by induction question
    Posted in the Discrete Math Forum
    Replies: 9
    Last Post: Dec 24th 2009, 02:47 PM
  5. Proof Question: Using Mathematical Induction
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Feb 19th 2009, 01:47 AM

/mathhelpforum @mathhelpforum