document.write( "Question 1116740: Please help me with this problem: Resolve into infinite series \"ax%2F%28a-x%29\"\r
\n" ); document.write( "\n" ); document.write( "I can't show how I tried on this problem because it is very complicated! I did try several times though.\r
\n" ); document.write( "\n" ); document.write( "This is the answer from my key book: x+x^2/a+x^3/a+x^4/a,ect., to infinity\r
\n" ); document.write( "\n" ); document.write( "I don't understand how the fractions are able to keep repeating. When I worked the problem I got as far as x+x^2/a, but no farther.
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Algebra.Com's Answer #731660 by ikleyn(52780)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "            The formula in your post is written   I N C O R R E C T L Y.\r
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\n" ); document.write( "\n" ); document.write( "            The correct formula is   T H I S:\r
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\n" ); document.write( "\n" ); document.write( "            \"ax%2F%28a-x%29\" = \"x\" + \"x%5E2%2Fa\" + \"x%5E3%2Fa%5E2\" + \"x%5E4%2Fa%5E3\" +  .  .  .  ,   ect.,   to infinity.\r
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\n" ); document.write( "It is not difficult to prove it   //   if you know the formula for the sum of an infinite geometric series . . . \r
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document.write( "\"a%2F%28a-x%29\" = \"a%2A%281%2F%28a%2A%281-x%2Fa%29%29%29%29\" = \"1%2F%281-x%2Fa%29\".\r\n" );
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document.write( "The very right fraction  \"1%2F%281-x%2Fa%29\"  is the sum of an infinite geometric progression with the first term of \"1\" and the common ratio of \"x%2Fa\":\r\n" );
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document.write( "\"1%2F%281-x%2Fa%29\" =  1 + \"x%2Fa\" + \"x%5E2%2Fa%5E2\" + \"x%5E3%2Fa%5E3\" + . . .      (1)\r\n" );
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document.write( "Now multiply both sides of (1) by x, and you will get the required solution\r\n" );
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document.write( "\"%28ax%29%2F%28a-x%29\" = x + \"x%5E2%2Fa\" + \"x%5E3%2Fa%5E2\" + \"x%5E4%2Fa%5E3\" + . . .\r\n" );
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document.write( "QED.\r\n" );
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