document.write( "Question 65740: Another problem I need a little help with.
\n" ); document.write( "Find the sum of the infinite geometric series if possible.
\n" ); document.write( "-1+(1/4)-(1/16)+(1/64)....
\n" ); document.write( "I multiplied by 4 and came up with 4s=-4+1-(1/4)+(1/16). What do I do next? Or maybe it's not possible to find the sum.....???
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Algebra.Com's Answer #46464 by praseenakos@yahoo.com(507)\"\" \"About 
You can put this solution on YOUR website!
QUESTION:\r
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\n" ); document.write( "\n" ); document.write( "Find the sum of the infinite geometric series if possible.
\n" ); document.write( "-1+(1/4)-(1/16)+(1/64)....\r
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\n" ); document.write( "\n" ); document.write( "ANSWER:\r
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\n" ); document.write( "\n" ); document.write( "Here the series is, -1+(1/4)-(1/16)+(1/64)....\r
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\n" ); document.write( "\n" ); document.write( "We can see that first term is -1 and common ratio is -1/4
\n" ); document.write( "(you can find out common ratio of this series by dividing second term by third term)\r
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\n" ); document.write( "\n" ); document.write( "Sum to infinity of a geometric seriesn is given by the formula,\r
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\n" ); document.write( "\n" ); document.write( "S = a/(1-r) where 'a' is the first term and 'r' is the common ratio.\r
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\n" ); document.write( "\n" ); document.write( "So we have,\r
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\n" ); document.write( "\n" ); document.write( "S = -1/( 1-{-1/4} )\r
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\n" ); document.write( "\n" ); document.write( "==> S = -1/( 1+{1/4} )\r
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\n" ); document.write( "\n" ); document.write( "==> S = -1/( 1*4 +1)/4\r
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\n" ); document.write( "\n" ); document.write( "==> S = -1/ (5/4)
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\n" ); document.write( "\n" ); document.write( "==> S = -4/5\r
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\n" ); document.write( "\n" ); document.write( "Which is the required answer.\r
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\n" ); document.write( "\n" ); document.write( "Hope you understood.\r
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\n" ); document.write( "\n" ); document.write( "Regards.\r
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\n" ); document.write( "\n" ); document.write( "Praseena.\r
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