document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=618022>Question 618022</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Use induction to prove that 1*2+2*3+3*4+...+n(n+1)=(n(n+1)(n+2))/3 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #388904 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=388904\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> n=1 base case is true. Assume it is true for some n = k. Show that it is true for k+1. In other words, show that\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE 1*2 + 2*3 + ... + k(k+1) + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}\">\r<BR>\n" );
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document.write( "Hint: What is 1*2 + ... + k(k+1)? </SPAN>\n" );
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