document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=250953>Question 250953</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find \r<BR>\n" );
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document.write( "-4+((100*(pi^2))/2!)-((2500*(pi^4))/4!)+((62500*(pi^6))/6!)-...<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #184455 by </B><A HREF=/tutors/aboutme.mpl?userid=drk><B>drk(1908)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=drk><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=184455\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> We can create a formula for the alternating sum. It might look a little messy:\r<BR>\n" );
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document.write( "SUM(n=0 to inf.) (-1)^n^2*pi^(2n+2)  /  (2n+2)factorial(!)\r<BR>\n" );
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document.write( "If we sum all of this, we get 8.\r<BR>\n" );
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document.write( "Now, -4 + 8 = 4.<BR>\n" );
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