document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=959118>Question 959118</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Write the series with summation notation: 2+5+8+11+... </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #586211 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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=586211\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Method #1\r<BR>\n" );
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document.write( "Method #2\r<BR>\n" );
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document.write( "Notice how we add 3 to each term to get the next term (2+3 = 5, 5+3 = 8, etc). This points to the fact that we have an arithmetic sequence. \r<BR>\n" );
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document.write( "So the common difference is d = 3\r<BR>\n" );
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document.write( "The first term is a(1) = 2\r<BR>\n" );
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document.write( "Use these two facts to get the nth term\r<BR>\n" );
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document.write( "a(n) = a(1) + d(n-1)<BR>\n" );
document.write( "a(n) = 2 + 3(n-1)<BR>\n" );
document.write( "a(n) = 2 + 3n-3<BR>\n" );
document.write( "a(n) = 3n - 1\r<BR>\n" );
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document.write( "So because the nth term is 3n-1 this means that <IMG STYLE=\"max-width:80%;\" src=\"http://www.sciweavers.org/tex2img.php?eq=2%2B5%2B8%2B11%2B...%20%3D%20%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%283k-1%29&bc=White&fc=Black&im=jpg&fs=12&ff=arev&edit=0\" align=\"center\" border=\"0\" alt=\"2+5+8+11+... = \sum_{k=1}^{\infty}(3k-1)\" width=\"269\" height=\"50\" /> (the only difference is that I'm using k instead of n as the index)\r<BR>\n" );
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document.write( "Jim\r<BR>\n" );
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