document.write( "Question 871381: A recursive formula for the following series: 1, 3/2, 5/6, 7/24, 9/120 \n" ); document.write( "

Algebra.Com's Answer #525572 by KMST(5328) $\"\"$ $\"About$

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A recursive formula shows how to get the next term from the one before, to $\"X%5Bn%5D\"$ from $\"X%5Bn-1%5D\"$ .
\n" ); document.write( "I know the idea to get $\"x%5Bn%5D\"$ from $\"X%5Bn-1%5D\"$ .
\n" ); document.write( "From the terms you are given, in the form given,
\n" ); document.write( "to get term number $\"n\"$ ,
\n" ); document.write( "you add 2 to the numerator and multiply the denominator by $\"n\"$ .
\n" ); document.write( "So for the 6th term, you take $\"9%2F120\"$ ;
\n" ); document.write( "change the numerator from $\"9\"$ to $\"9%2B2=11\"$ ,
\n" ); document.write( "and change the denominator from $\"120\"$ to $\"120%2A6=720\"$ ,
\n" ); document.write( "to get $\"11%2F720\"$ .
\n" ); document.write( "
\n" ); document.write( "However, if I was given $\"3%2F40\"$ as a term, how would I know that $\"3%2F40=9%2F120\"$ ,
\n" ); document.write( "and how would I know that $\"3%2F40=9%2F120\"$ is the fifth term.
\n" ); document.write( "It is not so easy.
\n" ); document.write( "I could do it, but I do not know how to write it as a recursive formula..
\n" ); document.write( "
\n" ); document.write( "I do see an explicit formula to calculate term number $\"n\"$ from $\"n\"$ , not counting on term number $\"n-1\"$ ) .
\n" ); document.write( " $\"X%5Bn%5D=%282n-1%29%2Fn%21\"$
\n" ); document.write( " $\"X%5B1%5D=%282%2A1-1%29%2F1%21=%282-1%29%2F1=1%2F1=1\"$
\n" ); document.write( " $\"X%5B2%5D=%282%2A2-1%29%2F2%21=%284-1%29%2F%281%2A2%29=3%2F2\"$
\n" ); document.write( " $\"X%5B3%5D=%282%2A3-1%29%2F3%21=%286-1%29%2F%281%2A2%2A3%29=5%2F6\"$
\n" ); document.write( " $\"X%5B4%5D=%282%2A4-1%29%2F4%21=%288-1%29%2F%281%2A2%2A3%2A4%29=7%2F24\"$
\n" ); document.write( " $\"X%5B5%5D=%282%2A5-1%29%2F5%21=%2810-1%29%2F%281%2A2%2A3%2A4%2A5%29=9%2F120\"$
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\n" ); document.write( "If you only told me that one of the terms is $\"3%2F40\"$ ,
\n" ); document.write( "I would factor $\"40\"$ as $\"40=2%2A2%2A2%2A5=2%2A4%2A5\"$ .
\n" ); document.write( "I would multiply numerator and denominator times $\"3\"$ ,
\n" ); document.write( "to have $\"2%2A3%2A4%2A5=5%21\"$ as a denominator.
\n" ); document.write( "Then, suspecting that $\"3%2F40=3%2A3%2F%2840%2A3%29=9%2F120\"$ is term number $\"n=5\"$ ,
\n" ); document.write( "I would verify that with $\"n%21=5%21\"$ as denominator, the numerator is $\"2n-1=2%2A5-1=10-1=9\"$ .
\n" ); document.write( "After that, from $\"9%2F120\"$ ,
\n" ); document.write( "I would change the numerator from $\"9\"$ to $\"9%2B2=11\"$ ,
\n" ); document.write( "and change the denominator from $\"120\"$ to $\"120%2A6=720\"$ ,
\n" ); document.write( "to get $\"11%2F720\"$ .
\n" ); document.write( "I cannot write that as a recursive formula,
\n" ); document.write( "but since no one else answered your question before,
\n" ); document.write( "I assume that not many people can.
\n" ); document.write( " \n" ); document.write( "

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