$\displaystyle \sum ^m _{j=1} 2^j-2^{j-1}$ $\displaystyle [2^1-2^{m+1}][1-2]^{-1}-[2^1-2^m][1-2]^{-1}$ $\displaystyle 2^{m+1}-2-2^m+2$ $\displaystyle 2^m$ Can someone pin-point where I went wrong?
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Originally Posted by quantoembryo $\displaystyle \sum ^m _{j=1} 2^j-2^{j-1}$ $\displaystyle [2^1-2^{m+1}][1-2]^{-1}-[2^1-2^m][1-2]^{-1}$ $\displaystyle 2^{m+1}-2-2^m-2$ $\displaystyle 2^m$ Can someone pin-point where I went wrong? $\displaystyle 2^j-2^{j-1}=2^{j-1} (2-1)=2^{j-1}$
Okay, but why cannot one merely apply the summation formula from the very beggining, like I had done?
Originally Posted by quantoembryo Okay, but why cannot one merely apply the summation formula from the very beggining, like I had done? $\displaystyle \sum ^m _{j=1} 2^j-2^{j-1}=\sum ^m _{j=1}2^j+\sum ^m _{j=1}2^{j-1}=...$
Whoops, just realized I made a technical error. I see where I went wrong. Thanks!
Just a side note: this sum telescopes.
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