document.write( "Question 1084210: Evaluate the sum:\r
\n" ); document.write( "\n" ); document.write( "4/sigma/n=1 (2/3)^n-1
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Algebra.Com's Answer #698282 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "I'm assuming the summation you want to do is
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\n" ); document.write( "\n" ); document.write( "If n = 1, then (2/3)^(n-1) = (2/3)^(1-1) = (2/3)^0 = 1
\n" ); document.write( "If n = 2, then (2/3)^(n-1) = (2/3)^(2-1) = (2/3)^1 = 2/3
\n" ); document.write( "If n = 3, then (2/3)^(n-1) = (2/3)^(3-1) = (2/3)^2 = 4/9
\n" ); document.write( "If n = 4, then (2/3)^(n-1) = (2/3)^(4-1) = (2/3)^3 = 8/27\r
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\n" ); document.write( "\n" ); document.write( "Therefore, summing from n = 1 to n = 4 yields\r
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\n" ); document.write( "\n" ); document.write( "The answer would be the fraction 65/27.\r
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