# Find the first 5 terms

• Mar 30th 2007, 03:27 PM
arturju
Find the first 5 terms
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!

• Mar 30th 2007, 03:49 PM
Plato
What have you done?
Where do you have a problem?
There are very straight forward!
• Mar 31st 2007, 03:40 AM
arturju
uhm.. for a. what does it mean by 'first 5 terms?' is that n+1, n+2.. n+5?
for b. what are the first 5 terms...? anf for c. also i dont know what the first 5 terms are
• Mar 31st 2007, 05:45 AM
topsquark
Quote:

Originally Posted by arturju
Find the first five terms of the recursively defined sequences bellow,

a. a1 = 1, an+1=1 + 1/n

Quote:

Originally Posted by arturju
uhm.. for a. what does it mean by 'first 5 terms?' is that n+1, n+2.. n+5?
for b. what are the first 5 terms...? anf for c. also i dont know what the first 5 terms are

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
• Mar 31st 2007, 04:55 PM
arturju
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

• Mar 31st 2007, 05:00 PM
Jhevon
Quote:

Originally Posted by arturju
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

i only examined c since you said you weren't sure about that. it's correct. you can post the next problem in a new thread
• Mar 31st 2007, 05:02 PM
Jhevon
b looks ok too