document.write( "Question 1144530: find sum to infinity for following sequence
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Algebra.Com's Answer #765681 by Edwin McCravy(20056) $\"\"$ $\"About$

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1 + 2/3 + 3/9 + 4/27 + 5/81 + ...
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document.write( " 1/1 + 2/3 + 3/9 + 4/27 + 5/81 +  ...\r\n" );
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document.write( "The sequence of numerators is 1, 2, 3, 4, 5. ..., which has nth term n.\r\n" );
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document.write( "The sequence of denominators is 1, 3, 9, 27, 81, ..., which has nth term 3ⁿ.\r\n" );
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document.write( "So the series we want has nth term \r\n" );
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document.write( "We want the infinite sum \r\n" );
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document.write( "We start with the geometric series formula\r\n" );
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document.write( "Substitute a=1\r\n" );
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document.write( "Differentiate both sides with respect to r:\r\n" );
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document.write( "Substitute r=1/3\r\n" );
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document.write( "We need to make the coefficient match the exponent. \r\n" );
document.write( "So we write the first n as n-1+1 and then as (n-1)+1\r\n" );
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document.write( "Distribute\r\n" );
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document.write( "Distribute the sum:\r\n" );
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document.write( "We change the limits, so that n-1 will be a single letter.\r\n" );
document.write( "We want n-1 = k, so we substitute n = k+1 \r\n" );
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document.write( "Now we can go back to our original notation by changing k to n:\r\n" );
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document.write( "But we now have n starting at -1.  But we want n to start at 0, so we write\r\n" );
document.write( "out the first term separately, then the rest of the sum will begin at n=0\r\n" );
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document.write( "Simplifying, the terms we wrote separately are -3 and 3, so they cancel:\r\n" );
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document.write( "The second sum is an infinite geometric series with a=1 an r=1/3 so we use\r\n" );
document.write( "the sum formula for an infinite geometric series and get \r\n" );
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document.write( "       \r\n" );
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document.write( "So we substitute 3/2 for the second sum and have:\r\n" );
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document.write( "and solve for the sum, which is what we want:\r\n" );
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document.write( "Edwin

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