document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=872382>Question 872382</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I just had a simple question.<BR>\n" );
document.write( "This is really going to sound pathetic and its really like grade 1 maths but for the life of me I cannot figure out the proof of this formula.Here is the proof:<BR>\n" );
document.write( "Geometric Progression = a + ar + ar^2 +  ar^3 + ... + ar^n-1  (1)<BR>\n" );
document.write( "multiply formula (1) by r:    r.GP = ar + ar^2 +  ar^3 + ... + ar^n-1 + ar^n (2)<BR>\n" );
document.write( "subtracting formula (2) from (1)gives: GP-  r.GP = a -ar^n<BR>\n" );
document.write( "factoring on both sides gives:  GP(1-r)=a(1-r^n)<BR>\n" );
document.write( "dividing by (1-r): GP= a(1 -r^n)/ (1-r)<BR>\n" );
document.write( "My problem in the proof is not the logic  behind it its the mechanics of the maths in line 2 of the proof where formula 1 gets multiplied through by r, for the life of me I dont understand why by multiplying through by \"r\" it leaves the term \"ar^n-1\". I though by multiplying the term ar^n-1 in the first formula it leaves ar^n due to the laws of exponents thus I do not see where the original term comes from in the second line. Its probably one of the most fundamental laws of algebra and I feel really bad not being able to get it, I just would really like some help on it please.<BR>\n" );
document.write( "P.S. I got the proof from this url  http://www.mathcentre.ac.uk/resources /workbooks/mathcentre/APGP.pdf page 9 at the bottom. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #526106 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=526106\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Geometric Progression = GP  = <BR>\n" );
document.write( "a + ar + ar^2 + ar^3 + ... + ar^(n-2) + ar^n-1  (1)<BR>\n" );
document.write( "multiply formula (1) by r to get: <BR>\n" );
document.write( "r.GP = ar + ar^2 + ar^3 + ... + ar^(n-1) + ar^n (2)<BR>\n" );
document.write( "-----------------------------------------------------------<BR>\n" );
document.write( "subtracting formula (2) from (1) gives: GP- r.GP = a -ar^n<BR>\n" );
document.write( "factoring on both sides gives: GP(1-r)=a(1-r^n)<BR>\n" );
document.write( "dividing by (1-r): GP= a(1 -r^n)/ (1-r)<BR>\n" );
document.write( "-----------------------------------------------<BR>\n" );
document.write( "My problem in the proof is not the logic behind it its the mechanics of the maths in line 2 of the proof where formula 1 gets multiplied through by r, for the life of me I dont understand why by multiplying through by \"r\" it leaves the term \"ar^n-1\". <BR>\n" );
document.write( "---<BR>\n" );
document.write( "Ans: ar^(n-1) is the result of multiplying ar^(n-2) by r<BR>\n" );
document.write( "r[ar^(n-2) = ar^(n-2+1) = ar^(n-1)<BR>\n" );
document.write( "----<BR>\n" );
document.write( "Each term of (2) simply adds \"1\" to the exponent of \"r\"<BR>\n" );
document.write( "in (1).<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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document.write( " </SPAN>\n" );
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