document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=38086>Question 38086</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Domenica invests $200 each quarter in a fixed interest mutualfund paying an annual interest of 7% compoundedquarterly. How much will her account will her account have in it at the end of ten yrs? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #23589 by </B><A HREF=/tutors/aboutme.mpl?userid=josmiceli><B>josmiceli(19441)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=josmiceli><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=josmiceli><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=23589\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Not easy! I copied this from an interest calculator website. If you compound<BR>\n" );
document.write( "quarterly AND make quarterly contributions, it gets messy. The key is not<BR>\n" );
document.write( "confusing your terms. Know what they mean.<BR>\n" );
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document.write( "One thing we need to get straight from the beginning is the timing of the interest and contributions. We're going to assume that the balance of the account for any particular year includes the interest growth from last year, but does not yet include the new contribution. (This is the way most people do it, but sometimes you'll see an example that assumes a different schedule; they will disagree with our formula by the equivalent of one year's interest and/or one year's contribution).\r<BR>\n" );
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document.write( "We'll write c for the annual contribution; and to keep things cleaner we'll write \"z\" for (1 + r). Now we start writing down the account balance for the first few years:\r<BR>\n" );
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document.write( "Year Balance<BR>\n" );
document.write( "Now P<BR>\n" );
document.write( "1 (P + c)z<BR>\n" );
document.write( "2 ((P + c)z + c)z\r<BR>\n" );
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document.write( "In other words, to go from one year's balance to the next, you add on the contribution c, then multiply by z to get the interest.\r<BR>\n" );
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document.write( "If you multiply these terms out, you'll start to see the pattern emerge:\r<BR>\n" );
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document.write( "Year Balance<BR>\n" );
document.write( "Now P<BR>\n" );
document.write( "1 Pz + cz<BR>\n" );
document.write( "2 Pz2 + c(z + z2)<BR>\n" );
document.write( "n Pzn + c(z + z2 + . . . + zn)\r<BR>\n" );
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document.write( "The second part of that last line is just c times the sum of a geometric series. So you can boil the whole thing down to:\r<BR>\n" );
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document.write( "1. Balance(n) = Pzn + c[(zn + 1 - z)/(z - 1)]\r<BR>\n" );
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document.write( "Finally, write z out in terms of r, to get the formula we're looking for:\r<BR>\n" );
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document.write( "2. Balance(n) = P(1 + r)n + c[((1 + r)n + 1 - (1 + r))/r]<BR>\n" );
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document.write( "P = what Dominica started out with. In this problem, it's $200, the same<BR>\n" );
document.write( "    as her quarterly contribution, since the problem didn't say different.<BR>\n" );
document.write( "c = her quarterly contribution of $200.<BR>\n" );
document.write( "n = the number of times the principle gets compounded. In the problem,<BR>\n" );
document.write( "    that would be 4 x 10 = 40, because quarterly is 4 times a year and the<BR>\n" );
document.write( "    total time is ten years.<BR>\n" );
document.write( "r = the compound interest rate of 7%<BR>\n" );
document.write( "Now the easy part<BR>\n" );
document.write( "Balance(n) = P(1 + r)n + c[((1 + r)n + 1 - (1 + r))/r]<BR>\n" );
document.write( "B = 200*(1 + .07)*40 + 200(((1 + .07)*40 + 1 - (1 + .07))/.07)<BR>\n" );
document.write( "B = 200*1.07*40 + 200((1.07*40 + 1 - 1.07)/.07)<BR>\n" );
document.write( "B = 8560 + 200(42.8 - .07)/.07)<BR>\n" );
document.write( "B = 8560 + (200 * 42.73)/.07<BR>\n" );
document.write( "B = 8560 + 8546/.07<BR>\n" );
document.write( "B = 8560 + 122086<BR>\n" );
document.write( "B = 130646 (rounded to nearest dollar)<BR>\n" );
document.write( "That's the answer unlass I made a mistake, which is entirely possible </SPAN>\n" );
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