1. ## Formula for N

Hiya

In the number pathern 1.3.6.10.15.21.34
I found the recursive formula an+1-an=n+1
now i need to find a formula for n.
shouldent it be somthing like an+1=n ?

2. Originally Posted by JBswe
Hiya

In the number pathern 1.3.6.10.15.21.34
I found the recursive formula an+1-an=n+1
now i need to find a formula for n.
shouldent it be somthing like an+1=n ?
why wouldn't it be $n = a_{n + 1} - a_n - 1$?

because n is actually not equal to a_{n + 1}, you can just look at the numbers and see that

3. 1, 3, 6, 10, 15, 21, 28, 36........

Is an easy sequence.

I take it you want to find an expression for the nth term?

As a clue notice..
1=1x2/2
3=2x3/2
6=3x4/2
10=4x5/2
etc.....

The one you posted 1,3,6,10,15,21,34 is more difficult.

4. Originally Posted by a tutor
1, 3, 6, 10, 15, 21, 28, 36........

Is an easy sequence.

[snip]

The one you posted 1,3,6,10,15,21,34 is more difficult.
Well spotted. At this thread (which I assume post #1 is following up from) that's the sequence I had in mind too ..... 1,3,6,10,15,21,34 is more difficult and the formula I gave, $a_{n+1} - a_n = n+1$, clearly fails to give $a_7 = 34$ ....

5. am sorry i messed up it is 1, 3, 6, 10, 15, 21, 28, not 34 as the 7th number

6. ok i did some recounting and the formula for n i think is an=n*(n-1/2) is this right?

7. Originally Posted by JBswe
ok i did some recounting and the formula for n i think is an=n*(n-1/2) is this right?

No. Even if you meant n(n-1)/2

This gives 0,1,3,6,10....

8. Originally Posted by a tutor
1, 3, 6, 10, 15, 21, 28, 36........

Is an easy sequence.

I take it you want to find an expression for the nth term?

As a clue notice..
1=1x2/2
3=2x3/2
6=3x4/2
10=4x5/2
etc.....

The one you posted 1,3,6,10,15,21,34 is more difficult.
A further clue is needed perhaps ....? Here's a massive one -

1=1x(1+1)/2
3=2x(2+1)/2
6=3x(3+1)/2
10=4x(4+1)/2
etc.....

The rule should come as no surprise since $a_n$ is the sum of the first n positive integers ......

btw 1 (as I suggested in the other thread) you could always look into the general method of solving a recurrence relation of the form $a_{n+1} - a_n = \, \text{polynomial function of } n$.

btw 2 it's been said that mathematics is the study of patterns. And what better way of developing this than questions like this one ......

9. ok think i got it

n=n(n+1)/2
and the recursive formula is
an+1-an=n+1

10. Originally Posted by JBswe
ok think i got it

$a_n$ = n(n+1)/2 Mr F correction of typo(?)
and the recursive formula is
an+1-an=n+1
Well you got there in the end. Thanks all round, I hope.

11. Ya now that i know i got it down a big thanks to all of you who help me out, Thank the gods you are here