Find the first five terms of the recursively defined sequences bellow,
a. a1 = 1, an+1=1 + 1/n
b. a1=2, a2=3,a3= 4, an-1 + an-3
c. a1=2, an+1=an!
Thx in advance!
I'll do the first one as an example and I would recommend you do the other two and post them. That way we can check your work and correct any mistakes you may have made.
a) a_1 = 1, a_{n+1} = 1 + 1/n
In general this series is defined as the set of numbers:
a_1, a_2, a_3, ...
The first 5 terms of the series would then be:
a_1, a_2, a_3, a_4, and a_5.
By the definition above, the first term, a_1, is given by
a_1 = 1.
Again, by the definition above, a_2 is given by the n = 2 term a_n = 1 + 1/n for n = 2. So:
a_2 = 1 + 1/2 = 3/2
Similarly
a_3 = 1 + 1/3 = 4/3
a_4 = 1 + 1/4 = 5/4
a_5 = 1+ 1/5 = 6/5
So the first five terms of the series are:
1, 3/2, 4/3, 5/4, and 6/5.
You should consult your textbook to make sure you understand what the different words mean (such as what a "term" of a series is) to help you out.
-Dan
if the first five terms for the 1st one are as you said, then for the 2nd one wouldnt i have tu find just the next one?
Nevertheless, i worked on the problem and i got this:
b. a1=2, a2=3,a3= 4, an-1 + an-3
~~~~~~~~~~~~~~~~~~~~~
a4=a4-1+a4-3=4+2=6
a5=a5-1+a5-3=6+3=9
a6=a6-1+a6-3=9+4=13
a7=a7-1+a7-3=13+6=19
...and so on..
I think i did problem c. wrong but i'll post it here anyways:
c. a1=2, an+1=an!
~~~~~~~~~~~~~~~~~~~~~
a2=a1+1=a1!=2
a3=a2+1=a2!=2
... and so i got 2 for all of them..
I have another problem like this i'm having trouble with but i'd rather post it after i get this one over with. Thx in advance