Series Help

[mageslayer]

1, 4, 10, 19, 31, ...

I need to write a formula to generate the nth term.

I believe it is a geometric sequence, however, I cannot make a formula.

 

[Gibsons]

X = 1
then a loop with something sort of kinda like
X = X + n
n=n+3

dunno if you want some sort of algorithm, but it shouldn't be too hard, just add 3 more each time. Next number should be 46...

 

[mageslayer]

I can see the pattern and find the following numbers, but I cannot make a formula for the nth term in the sequence.

 

[CycloWizard]

y_n+1=y_n+3n

 

[mageslayer]

Originally posted by: CycloWizard
y_n+1=y_n+3n

that doesn't seem to work for the first term

y_1=y_0+3(0)

=0 ???

 

[CycloWizard]

Originally posted by: mageslayer
that doesn't seem to work for the first term

y_1=y_0+3(0)

=0 ???

The entire thing is just the successive terms of the summation y=1+sum_n=1^i{3n}. What I gave previously was the recursive relationship that occurs within the sum. I didn't think I'd have to do ALL of your homework for you. :roll:

 

[esun]

http://www.research.att.com/~n...uage=english&go=Search

 

[Rasico]

The answer I believe is:
3 * (n)(n-1)/2 + 1

This is assuming the answer is 1 for n = 1.

Everyone pointed out the obvious: f(1) = 1
then f(n) = f(n-1) + 3*n.

So lets look for a pattern:
like f(f)

f(3) = f(2) + 3 * 3
= f(1) + 3*2 + 3*3

If we generlize this with n we get

f(n) = f(n-1) + 3n = f(n-2) + 3(n-1) + 3*(n-2) which would be the sum of all numbers from n to n-1 x 3 + 1.

The sum of all numbers from 1 to n-1 is n*(n-1)/2. Hope that made sense.