document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209828>Question 1209828</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The Fibonacci sequence, is defined by F_0 = 0, F_1 = 1, and F_n = F_{n - 2} + F_{n - 1}.  It turns out that<BR>\n" );
document.write( "F_n = \frac{\alpha^n - \beta^n}{\sqrt{5}},<BR>\n" );
document.write( "where \alpha = \frac{1 + \sqrt{5}}{2} and \beta = \frac{1 - \sqrt{5}}{2}.\r<BR>\n" );
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document.write( "The Lucas sequence is defined as follows: L_0 = 2, L_1 = 1, and<BR>\n" );
document.write( "L_n = L_{n - 1} + L_{n - 2}<BR>\n" );
document.write( "for n \ge 2.  What is L_4?<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 #850737 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=850737\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's calculate $L_4$ using the given recurrence relation and initial values.\r<BR>\n" );
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document.write( "* $L_0 = 2$<BR>\n" );
document.write( "* $L_1 = 1$<BR>\n" );
document.write( "* $L_n = L_{n-1} + L_{n-2}$ for $n \ge 2$\r<BR>\n" );
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document.write( "Now, let's find the subsequent terms:\r<BR>\n" );
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document.write( "* $L_2 = L_1 + L_0 = 1 + 2 = 3$<BR>\n" );
document.write( "* $L_3 = L_2 + L_1 = 3 + 1 = 4$<BR>\n" );
document.write( "* $L_4 = L_3 + L_2 = 4 + 3 = 7$\r<BR>\n" );
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document.write( "Therefore, $L_4 = 7$.\r<BR>\n" );
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document.write( "Final Answer: The final answer is $\boxed{7}$<BR>\n" );
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