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Just one line is giving me trouble:

$(2^{-l+1} + 2^{-l} + 2^{-l-1}+ \cdots + 2^{-m+2} )\cdot M$

$=[(2^{-l+2} - 2^{-l+1}) + (2^{-l-1} - 2^{-l}) + \cdots +(2^{-m+3} - 2^{-m+2}) ]\cdot M$

Hows that?

  • 1
    $\,2^n+2^n=2\cdot 2^n=2^{n+1}\,$2012-09-25

2 Answers 2

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Its from the fact that $2^{k+1} - 2^k = 2^k(2-1) = 2^k$ Therefore each power of $2$ is expanded as a difference.

i.e. $2^{-l+1} = 2^{-l+2} - 2^{-l+1}$ and so on....

2

Oh, your comment made me see! Check that for any $\,n\in\Bbb N\,$:

$2^n-2^{n-1}=2^{n-1}\left(2-1\right)=2^{n-1}...!!$

so $\,2^{-l+1}=2^{-l+2}-2^{-l+1}\,$ and etc.