Question 871381
A recursive formula shows how to get the next term from the one before, to {{{X[n]}}} from {{{X[n-1]}}} .
I know the idea to get {{{x[n]}}} from {{{X[n-1]}}} .
From the terms you are given, in the form given,
to get term number {{{n}}} ,
you add 2 to the numerator and multiply the denominator by {{{n}}} .
So for the 6th term, you take {{{9/120}}} ;
change the numerator from {{{9}}} to {{{9+2=11}}} ,
and change the denominator from {{{120}}} to {{{120*6=720}}},
to get {{{11/720}}}.
 
However, if I was given {{{3/40}}} as a term, how would I know that {{{3/40=9/120}}} ,
and how would I know that {{{3/40=9/120}}} is the fifth term.
It is not so easy.
I could do it, but I do not know how to write it as a recursive formula..
 
I do see an explicit formula to calculate term number {{{n}}} from {{{n}}} , not counting on term number {{{n-1}}} ) .
{{{X[n]=(2n-1)/n!}}}
{{{X[1]=(2*1-1)/1!=(2-1)/1=1/1=1}}}
{{{X[2]=(2*2-1)/2!=(4-1)/(1*2)=3/2}}}
{{{X[3]=(2*3-1)/3!=(6-1)/(1*2*3)=5/6}}}
{{{X[4]=(2*4-1)/4!=(8-1)/(1*2*3*4)=7/24}}}
{{{X[5]=(2*5-1)/5!=(10-1)/(1*2*3*4*5)=9/120}}}
 
If you only told me that one of the terms is {{{3/40}}} ,
I would factor {{{40}}} as {{{40=2*2*2*5=2*4*5}}} .
I would multiply numerator and denominator times {{{3}}} ,
to have {{{2*3*4*5=5!}}} as a denominator.
Then, suspecting that {{{3/40=3*3/(40*3)=9/120}}} is term number {{{n=5}}} ,
I would verify that with {{{n!=5!}}} as denominator, the numerator is {{{2n-1=2*5-1=10-1=9}}}.
After that, from  {{{9/120}}} ,
I would change the numerator from {{{9}}} to {{{9+2=11}}} ,
and change the denominator from {{{120}}} to {{{120*6=720}}},
to get {{{11/720}}}.
I cannot write that as a recursive formula,
but since no one else answered your question before,
I assume that not many people can.