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Math Help - simple proof by induction question

  1. #1
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    simple proof by induction question

    I have come across this 'simple' question in more than one book on proofs, yet I am not able to find a solution:

    Prove, for all n\geq0, that (1+x)^n\geq1+nx if 1+x>0.

    I can see how this fact can be demonstrated via binomial theorem

    (1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+...

    (since the first third component of the expansion is always non-negative)

    but could you give me a hint how to go about proving it using proof by induction?

    thanks
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Volga View Post
    I have come across this 'simple' question in more than one book on proofs, yet I am not able to find a solution:

    Prove, for all n\geq0, that (1+x)^n\geq1+nx if 1+x>0.

    I can see how this fact can be demonstrated via binomial theorem

    (1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+...

    (since the first third component of the expansion is always non-negative)

    but could you give me a hint how to go about proving it using proof by induction?

    thanks
    Merely note that \left(1+x)^{n+1}=(1+x)^n(1+x)\geqslant (1+nx)(1+x)=1+nx+x+nx^2=1+(n+1)x+nx^2\geqslant 1+(n+1)x.
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  3. #3
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    I would try

    For n=1: \displaystyle (1+x)^1 \geq 1+x True!

    Assume n=k: \displaystyle (1+x)^k \geq 1+kx is also true.

    Now for n=k+1:

    \displaystyle (1+x)^{k+1} \geq 1+(k+1)x

    \displaystyle (1+x)^{k}(1+x)^{1} \geq 1+kx+x

    Now use \displaystyle (1+x)^k \geq 1+kx to finish the argument.
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  4. #4
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    Oh thanks! I was stupid enough to use x=k and x=k+1 instead of n=k and n=k+1!!! Now it is obvious...
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