*Originally posted by joe shmo*

**would it be something like this for the above series
**

a(n) = [ a(n-1) - a(n-2) + 1] + a(n-1)

i see no way to move to a(n) without figuring out all previous a(n)'s

Is there

To explain a little more, one way to mathematically do it is by recursive substitution (this is the brute force, no-shortcuts method).

You have:

(1) a(n) = [ a(n-1) - a(n-2) + 1] + a(n-1) = 2.a(n-1) - a(n-2) + 1

But if this is valid for all n, then it is also valid for n-1:

(2) a(n-1) = 2.a(n-2) - a(n-3) + 1

Replace (2) for the a(n-1) (1) and you have:

a(n) = 4.a(n-2) - 2.a(n-3) + 2 - a(n-2) + 1 = 3. a(n-2) - 2.a(n-3) + 2 + 1

Do it again for a(n-2) as in (2), you have:

a(n-2) = 2.a(n-3) - a(n-4) + 1

=> a(n) = 4.a(n-3) - 3.a(n-4) + 3 + 2 + 1

Do you see a pattern here?

a(n) = (t+1)a(n-t) - t.a(n-t-1) + t + t-1 + ... + 1

Choose t = n-2 and you have:

a(n) = (n-1)a(2) - (n-2)a(1) + n-2 + n-3 + ... + 1

a(n) = 8n - 8 - 6n + 12 + n-2 + n-3 + ... + 1

a(n) = 2n + 4 + n-2 + n-3 + ... + 1

Since 2n = n + n-1 + 1, you can change the last term to:

a(n) = 5 + n + n-1 + n-2 + n-3 + ... + 1

Uf.

**Alternatively** (which was what I did). You can look at your series and see a pattern:

Value-> 6, 8, 11, 15, 20

Rank-> 1, 2, 3 , 4 , 5

Each number is the previous number plus its rank:

(3) a(n) = a(n-1) + n

(4) a(n-1) = a(n-2) + n-1

Substitute (4) in (3) and you have:

a(n) = a(n-2) + n + n-1

...

a(n) = a(1) + n + n-1 + ... + 2

Since a(1) = 6 = 5+1

a(n) = 5 + 1 + 2 + ... + n-1 + n