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How can I prove that the following equality holds only for $x=0$?

$$\binom{n}{2}x^2 +\cdots+ \binom{n}{n}x^n=0\text{ when }x\gt-1\text{ and }n\gt1$$

Michael Hardy
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frosh
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    $(1+x)^n = 1 + {n\choose 1}x + {n\choose 2} x^2 + \ldots + {n\choose n}x^n$ so you can reduce your equation down to $(1+x)^{n} = 1+nx$. – Winther May 14 '15 at 21:57
  • Hi @Winther, I didn't understand what you mean by your comment. – frosh May 14 '15 at 22:04
  • I just simply meant to point out a simpler way to write the equation you have written down (in case you did not know this), i.e. that that sum you wrote down $= 0$ is nothing but the equation $ (1+x)^n - 1 - nx = 0$. – Winther May 14 '15 at 22:06
  • From this simpler formulation it is easy to prove what you want. For example consider the function $f(x) = (1+x)^n - 1 - nx$. For this we have $f'(x) = n[(1+x)^{n-1} - 1]$. Now try to show that $f' < 0$ for $x\in [-1,0)$ and $f' > 0$ for $x>0$. If you can do this then this proves that $f$ has a minimum at $x=0$ where $f(0) = 0$ ergo $f(x) > 0$ for all other $x$. – Winther May 14 '15 at 22:10
  • If $f(x) = (1+x)^n - 1 - n x$, we have $f(0) = 0$ and $f'(x) = n ((1+x)^{n-1} - 1)$. Thus $f'(x) > 0$ if $x > 0$ and $< 0$ if $-1 < x < 0$. We conclude that $f(x)$ has no zeros for $x \ge -1$ other than $0$. – Robert Israel May 14 '15 at 22:10
  • I am not allowed to use differentiation. – frosh May 14 '15 at 22:15
  • Induction of $n$ also works. Assume $(1+x)^{n} > 1+nx$ and show that this implies $(1+x)^{n+1}> 1 + (n+1)x$. See for example this question which has an answer that is very close to what you are asking. – Winther May 14 '15 at 22:18
  • $(1+nx)(1+x)=1+x+nx+nx^2\ge1+(n+1)x$ – frosh May 14 '15 at 22:23
  • Observ the last inequality is strict if $x\neq 0$. – Bernard May 14 '15 at 22:35
  • Please check my answer. – frosh May 14 '15 at 22:44

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