document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=476628>Question 476628</A>: <I><SPAN STYLE=\"word-break: break-all;\"> prove that a perfact number can be written as a sum of (2^n)-1 consicutive numbers for some n.(please give me the proof) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #326964 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=326964\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Proving it assumes that all perfect numbers N can be written as\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE N = 2^{n-1} (2^n - 1)\">, in which this is equal to\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE N = \frac{(2^n - 1)(2^n)}{2} = 1 + 2 + 3 + ... + (2^n - 1)\">\r<BR>\n" );
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document.write( "hence we are done. So if mathematicians somehow find an odd perfect number this theorem might not be true. </SPAN>\n" );
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