document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=877695>Question 877695</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What is the formula for:\r<BR>\n" );
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document.write( "1+(1+2)+(1+2+3+4)+(1+2+3+4+5+6+7+8)+...\r<BR>\n" );
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document.write( "and how is it derived? I do know the formula btw (-: but I am not sure how to derive it. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #529488 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=529488\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Recall 1+2+...+n = n(n+1)/2. If we can find a general formula for <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large 1 \cdot 2 + 2 \cdot 3 + 4 \cdot 5 + 8 \cdot 9 + \ldots + 2^n \cdot 2^n + 1\">, we are essentially done.\r<BR>\n" );
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document.write( "Rewrite as <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large (1^2 + 2^2 + 4^2 + \ldots + (2^n)^2) + (1+2+4+ \ldots + 2^n)\">. The latter term is equal to <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large 2^{n+1} - 1\">. Note that the first sum is geometric with common ratio 4 (can be written as 1^2 + 4^2 + ... + 4^n) and is equal to <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large \frac{4^{n+1} - 1}{3}\">.\r<BR>\n" );
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document.write( "Therefore the sum is <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large \frac{\frac{4^{n+1} - 1}{3} + 2^{n+1} - 1}{2}\">, which I'll leave you to simplify. </SPAN>\n" );
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