document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=504679>Question 504679</A>: <I><SPAN STYLE=\"word-break: break-all;\"> proof of <BR>\n" );
document.write( "1+4+9+16+25+36+...+n^2=(n(n+1)(2*n+1))/6<BR>\n" );
document.write( "plz answer me soon<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 #339732 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=339732\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> It is easy to prove via induction; but more difficult to derive the formula. To prove it by induction, note that the base case n = 1 holds. Assume it holds for n=k, e.g.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE 1^2 + ... + k^2 = \frac{k(k+1)(2k+1)}{6}\">. Then we want to show it works for k+1:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE 1^2 + ... + (k+1)^2 = (1 + ... + k^2) + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE = \frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6} = \frac{(k+1)(k(2k+1) + 6(k+1))}{6}\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE = \frac{(k+1)(2k^2 + 7k + 6)}{6} = \frac{(k+1)(k+2)(2k+3)}{6}\">\r<BR>\n" );
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document.write( "which is what we would've gotten if we substituted k+1 into the formula. Hence the statement is true for all positive integers n. </SPAN>\n" );
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