# Thread: Need help finding the pattern in this sequence of numbers

1. ## Need help finding the pattern in this sequence of numbers

I wasn't sure where to place this problem in the forums. I hope this area is OK. I'm trying to find a pattern in this series of numbers. How can I find a simplified expression, where I evaluate it at my given n, and the proper f(n) is returned?

n=4, f(4)=6
n=5, f(5)=10
n=6, f(6)=15
n=7, f(7)=21

I found that the function can be represented by the sum $\sum_{k=1}^{n-1}{(n-k)}$.

So for n=4, f(4)=3+2+1
n=5, f(5)=4+3+2+1
...

Is there a way I can find a simplified expression? For example, the series
2,4,8,16 can be represented by n^2. I was curious if I can find a simplified expression for this pattern.

Thanks

Edit: This should probably be moved to the number theory forum. Sorry

2. $f(n) = 1 + 2 + 3 + . . . . . .+ (n-2) + (n-1) = \frac{n(n-1)}2$

3. Originally Posted by Fourier
I wasn't sure where to place this problem in the forums. I hope this area is OK. I'm trying to find a pattern in this series of numbers. How can I find a simplified expression, where I evaluate it at my given n, and the proper f(n) is returned?

n=4, f(4)=6
n=5, f(5)=10
n=6, f(6)=15
n=7, f(7)=21

I found that the function can be represented by the sum $\sum_{k=1}^{n-1}{(n-k)}$.

So for n=4, f(4)=3+2+1
n=5, f(5)=4+3+2+1
...

Is there a way I can find a simplified expression? For example, the series
2,4,8,16 can be represented by n^2. I was curious if I can find a simplified expression for this pattern.

Thanks

Edit: This should probably be moved to the number theory forum. Sorry
Look at the difference table:

Code:
6   10   15   21
4    5    6
1    1
That the second differences are constant tells you that the series can be
generated by a quadratic in $n$.

Now just fit a general quadratic $f(n)=an^2+bn+c$ to the data.

RonL

4. Originally Posted by Fourier
I wasn't sure where to place this problem in the forums. I hope this area is OK. I'm trying to find a pattern in this series of numbers. How can I find a simplified expression, where I evaluate it at my given n, and the proper f(n) is returned?

n=4, f(4)=6
n=5, f(5)=10
n=6, f(6)=15
n=7, f(7)=21

I found that the function can be represented by the sum $\sum_{k=1}^{n-1}{(n-k)}$.

So for n=4, f(4)=3+2+1
n=5, f(5)=4+3+2+1
...

Is there a way I can find a simplified expression? For example, the series
2,4,8,16 can be represented by n^2. I was curious if I can find a simplified expression for this pattern.

Thanks

Edit: This should probably be moved to the number theory forum. Sorry
Plus this is the well known triangle numbers whos generating function is $\frac{n(n+1)}{2}$