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I know how mathematical induction works and the generic algorithm of proving a statement by the Principle of Mathematical Induction, but I'm having trouble proving the base case for a particular problem in my textbook.

For every $n\in\Bbb{Z}^+$ where $x\neq1$
$$(1+x)(1+x^2)(1+x^4)...(1+x^{2n})=\frac{1-x^{2n+1}}{1-x}$$

My attempt for base case $n=1$:
$LHS: (1+x)(1+x^2)=1+x+x^2+x^3$
$RHS: \frac{1-x^3}{1-x}=\frac{(1-x)(1+x+x^2)}{1-x}=1+x+x^2$

Which is obviously not true for all $x\neq1$

What am I doing wrong?

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  • $\begingroup$ for what $n$ we are starting? $\endgroup$ Commented Jul 26, 2015 at 15:11
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    $\begingroup$ it's not $2n$ and $2n+1$ but $2^n$ and $2^{n+1}$ $\endgroup$ Commented Jul 26, 2015 at 15:11
  • $\begingroup$ @Dr.SonnhardGraubner: I might be wrong, but I think that the restriction on $n$ implies that we start at $n=1$ $\endgroup$ Commented Jul 26, 2015 at 15:17

2 Answers 2

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The claim is wrong. The correct statement is $$(1+x^{2^0})(1+x^{2^1})(1+x^{2^2})...(1+x^{2^n})=\frac{1-x^{2^{n+1}}}{1-x}$$

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The correct formula you should be proving is:

$$(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\frac{1-x^{2^{n+1}}}{1-x} $$

You either copied the problem wrong, or there is a mistake in the problem.

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    $\begingroup$ the terms on the LHS should be in the form $1 + x^{2^i}$ $\endgroup$ Commented Jul 26, 2015 at 15:13
  • $\begingroup$ @Nan Ty, fixed it. $\endgroup$ Commented Jul 26, 2015 at 15:17

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