document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1021450>Question 1021450</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let f_n be the nth Fibonacci number. Show that for every natural n<BR>\n" );
document.write( "f_1 + f_2 + . . . + f_n = f_(n+2) &#8722; 1. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #637172 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=637172\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You can show by induction on n.\r<BR>\n" );
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document.write( "When n = 1, we have <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large f_1 = f_3 - 1\">, which is true since 1 = 2-1.\r<BR>\n" );
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document.write( "For some <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large k \ge 1\">, assume by hypothesis that <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large f_1 + \ldots + f_k = f_{k+2} - 1\">. Adding <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large f_{k+1}\"> to both sides gives <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large f_1 + \ldots + f_{k+1} = f_{k+1} + f_{k+2} - 1 = f_{k+3} - 1\">, so the expression holds for k+1.\r<BR>\n" );
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document.write( "Therefore, the equation holds for all natural n. </SPAN>\n" );
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