# Thread: Finding Formula for a Number Sequence

1. ## Finding Formula for a Number Sequence

I need to find the formula for the general term in the given sequences:

---------And---------

I'm not sure how to approach this, I've looked for common differences, would it work if I used this: ?

Thank you, I appreciate your help!

2. the first one follows

$\frac{n}{(n-1)^2}~,n= 3,4,5,\dots$

the second

$\frac{1}{3n}~,n= 1,2,3,\dots$

3. Originally Posted by pickslides
the first one follows

$\frac{n}{(n-1)^2}~,n= 3,4,5,\dots$

the second

$\frac{1}{3n}~,n= 1,2,3,\dots$

Thank you for the first one!

As for the second, the first term corresponds to n=1, the second to n=2, etc. When n=1 your formula: n/(n-1)^2 actually divides by zero, and when n=2 it gives you 1.

So it just needed to be adjusted to (n+2)/(n+1)^2.

Thank you so much, you got me thinking on the right track.

4. Originally Posted by thebristolsound

As for the second, the first term corresponds to n=1, the second to n=2, etc. When n=1 your formula: n/(n-1)^2 actually divides by zero,
incorrect, the formula provided for the second sequence was

$\frac{1}{3n}~,n= 1,2,3,\dots$

For $n=1:~ \frac{1}{3\times 1}=\frac{1}{3}$ as required

Originally Posted by thebristolsound
and when n=2 it gives you 1.
Incorrect again.

For $n=2:~ \frac{1}{3\times 2}=\frac{1}{6}$ as required

5. Originally Posted by pickslides
incorrect, the formula provided for the second sequence was

$\frac{1}{3n}~,n= 1,2,3,\dots$

For $n=1:~ \frac{1}{3\times 1}=\frac{1}{3}$ as required

Incorrect again.

For $n=2:~ \frac{1}{3\times 2}=\frac{1}{6}$ as required
My bad, I misdirected you. I meant to thank you for the formula pertaining to the second sequence: However, as for the first sequence, I had to modify to in order for it to work when n=1 and n=2. When n=1 your formula actually divides by zero, and when n=2 it produced 1.

Sorry for the inconvenience!

6. Both notations are equivalent. The only difference is that his sequence starts with n=3 while yours starts with n=1. His formula does not divide by 0 since the first element is when n=3.

7. Originally Posted by thebristolsound
I had to modify to in order for it to work when n=1 and n=2. When n=1 your formula actually divides by zero,
You didn't have to modify, my sequence starts at n=3 as stated.