document.write( "Question 84140: Use the geometric sequence of numbers 1, 1/2, 1/4, 1/8,.....to find the following:\r
\n" ); document.write( "\n" ); document.write( "a) What is r, the ratio between 2 consecutive terms?
\n" ); document.write( " Answer: r=1/2 divided by 1,\r
\n" ); document.write( "\n" ); document.write( " which is 1/2(fraction)
\n" ); document.write( "Is this correct\r
\n" ); document.write( "\n" ); document.write( "b) Using the formula for the sum of the first n terms of a geometric series, what is the sum of the first 10 terms? Round your answer to 4 decimals.
\n" ); document.write( "show work\r
\n" ); document.write( "\n" ); document.write( "1st term is 1X1/2=1/2
\n" ); document.write( "2nd term is 1/2X1/2=1/4
\n" ); document.write( "3rd term is 1/4X1/2=1/8
\n" ); document.write( "4th term is 1/8X1/2=1/16
\n" ); document.write( "5th term is 1/16X1/2=1/32
\n" ); document.write( "6th term is 1/32X1/2=1/64
\n" ); document.write( "7th term is 1/64X1/2=1/128
\n" ); document.write( "8th term is 1/128X1/2=1/256
\n" ); document.write( "9th term is 1/256X1/2=1/512
\n" ); document.write( "10th term is 1/512X1/2=1,024\r
\n" ); document.write( "\n" ); document.write( "c)Using the formula for the sum of the first n terms of a geometric series, what is the sum of the first 12 terms? Round to 4 decimals.
\n" ); document.write( "show work\r
\n" ); document.write( "\n" ); document.write( "11th term is 1/1,024X1/2=2,048
\n" ); document.write( "12 term is 1/2,048X1/2=4,096\r
\n" ); document.write( "\n" ); document.write( "d) What observation can you make about these sums? In particular, what whole number does it appear that the sum will always be smaller than?
\n" ); document.write( "1
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Algebra.Com's Answer #60609 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
a)
\n" ); document.write( "The ratio r is the factor to get from term to term. So
\n" ); document.write( "r=nth term/(n-1) term
\n" ); document.write( "\"r=%281%2F8%29%2F%281%2F4%29=%284%2F8%29=1%2F2\"
\n" ); document.write( "\"r=1%2F2\"\r
\n" ); document.write( "\n" ); document.write( "So the ratio is \"r=1%2F2\" (you are correct)
\n" ); document.write( "The sequence is cut in half each term, so the sequence is \"%281%2F2%29%5En\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "b)
\n" ); document.write( "The sum of a geometric series is
\n" ); document.write( "\"S=a%281-r%5En%29%2F%281-r%29\"where a=1
\n" ); document.write( "\"S=%281-%281%2F2%29%5E10%29%2F%281-%281%2F2%29%29\"So plug in n=10 to find the sum of the first 10 partial sums
\n" ); document.write( "\"S=%281-1%2F1024%29%2F%281%2F2%29\"
\n" ); document.write( "\"S=2046%2F1024\"
\n" ); document.write( "So the sum of the first ten terms is \"2046%2F1024\" or 1.99805 approximately\r
\n" ); document.write( "\n" ); document.write( "c)
\n" ); document.write( "Use the same formula to find the sum of the 1st 12 terms
\n" ); document.write( "\"S=a%281-r%5En%29%2F%281-r%29\"where a=1
\n" ); document.write( "\"S=%281-%281%2F2%29%5E12%29%2F%281-%281%2F2%29%29\"So plug in n=12 to find the sum of the first 12 partial sums
\n" ); document.write( "\"S=%281-1%2F4096%29%2F%281%2F2%29\"
\n" ); document.write( "\"S=8190%2F4096\"
\n" ); document.write( "So the sum of the first twelve terms is \"8190%2F4096\" or 1.99951 approximately\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "d)
\n" ); document.write( "It appears that the sums are approaching a finite number of 2. This is because each term is getting smaller and smaller. This observation is justified by the fact that if \"abs%28r%29%3C1\" then the infinite series will approach a finite number. In other words
\n" ); document.write( "If \"abs%28r%29%3C1\" (the magnitude of r has to be less than 1) then,
\n" ); document.write( "\"S=a%2F%281-r%29\"Where S is the infinite series. So if we let a=1 and r=1/2 we get
\n" ); document.write( "\"S=1%2F%281-%281%2F2%29%29\"
\n" ); document.write( "\"S=1%2F%281%2F2%29\"
\n" ); document.write( "\"S=2\"So this verifies that our series approaches 2.
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