1, 3/2, 5/6, 7/24, 9/120
i cant find a suitable a(sub n) term for n greater than or equal to 1.
numerator is odd numbers and denominator is n times the previous denominator, but how do i express this?
I tried using an induction method and systems of linear equations with
an^2+bn+c
then,
1=a+b+c
3/2=4a+2b+c
5/6=9a+3b+c
but this only gives me the first three n terms (not surprising)
halp pls
that is, (-7/12)(n^2)+(9/4)n-(2/3) is a suitable formula when 1<n<3
i need a formula for all n greater than or equal to 1.
halp a brainlet. i tried four tutors and they cant solve
nvm i got its 2n-1/n!
im a brainlet srry
>>8481114
First term:
1/1
Second term
3/2
Third Term
5/6
Which is 2n-1/n! n E N
>>8481118
is 2n-1 another way of saying odd?
>>8481114
https://www.algebra.com/algebra/homework/Sequences-and-series/Sequences-and-series.faq.question.871381.html
>>8481210
2n-1 is always odd for natural n
>inb4 what about n=0
https://oeis.org/
>>8481227
>inb4 what about n=0
so -1 is a even number then? wtf
>>8481119
>Not specifying whether 0 E N or not.
>>8481119
retard
>>8481473
nice mee mee
>>8481114
Starting with n=1, the sequence is:
a_n = (2n - 1)/(n!), where n! is the factorial function, a function that takes the positive integers less than or equal to n, and multiplies them altogether.