- 19 Nov '07 22:49

would it be something like this for the above series*Originally posted by joe shmo***like**

series, 6,8,11,15,20

i have read a little bit about it

a(1) =6 a(2) = 8 a(3)= [a(2) - a(1)+1] + a(2)

a(4) = [a(3) - a(2) +1] + a(3)...........

how is a(n) composed?

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 - 19 Nov '07 23:19 / 1 edit

a(n) = a(n-1) + n = a(n-2) + n-1 + n = 5 + 1 + 2 + ... + n*Originally posted by joe shmo***like**

series, 6,8,11,15,20

i have read a little bit about it

a(1) =6 a(2) = 8 a(3)= [a(2) - a(1)+1] + a(2)

a(4) = [a(3) - a(2) +1] + a(3)...........

how is a(n) composed?

So if your n is 100 you have 5 + (1 + 2 + 3 + ... + 100)

If I understand your notation. - 19 Nov '07 23:42 / 1 edit

so this is a simplified version of what i have above?*Originally posted by Palynka***a(n) = a(n-1) + n = a(n-2) + n-1 + n = 5 + 1 + 2 + ... + n**

So if your n is 100 you have 5 + (1 + 2 + 3 + ... + 100)

If I understand your notation.

Your version looks like an arithmatic sequence, is it? could you use the nth term formula in this instance?

which is a(n) = a(1) + (n-1)d

or would the be the sum of the first n terms S(n)= n/2[a(1)+a(n)] - 19 Nov '07 23:50 / 1 edit

so S(n) = n/2[ a(1) + a(n) ] + 5 for the series 6,8,11,15.......*Originally posted by joe shmo***so this is a simplified version of what i have above?**

Your version looks like an arithmatic sequence, is it? could you use the nth term formula in this instance?

which is a(n) = a(1) + (n-1)d

or would the be the sum of the first n terms S(n)= n/2[a(1)+a(n)] - 19 Nov '07 23:57 / 2 edits

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

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 - 20 Nov '07 12:06

Look at the alternative if it looks intimidating.*Originally posted by joe shmo***thanks, I'll work on attempting to fully understand this for kicks, but I have to be honest it looks intimidating....**

haha..... thanks again

Each term is the sum of the previous one plus its rank, i.e. a(5) = 20 = a(4) + 5; a(4) = 11 + 4 and so on. - 20 Nov '07 17:37

what about these triangular numbers plus five, I pulled the series from the "what you have is ....." thread the creator posted a problem with these numbers*Originally posted by wolfgang59***a(n) = 5 +{n(n+1)/2}**

Its triangular numbers plus five

pictorially its;

00000

0

00

000

0000

00000

000000

etcetera - 20 Nov '07 23:45

In all honesty, your answer is sufficient. Take the Fibonacci sequence, for instance, which takes the form 1,1,2,3,5,8...*Originally posted by joe shmo***right.....but that quite easy to understand so i will try to put some effort into the theory.**

Normally you would write it as a recurrence relation,

f[n]=f[n-2]+f[n-1].

However, we can also write it as

f[n]=((A^n)-(1-A)^n)/sqrt(5)

where A=(1+sqrt(5))/2. Writing it in any other form would be unnecessary. ~~chessisvanity~~THE BISHOP GOD- 21 Nov '07 00:11

thanks for that, I was just going to ask about the Fibonacci sequence and I had a hunch that the golden ratio could be used to figure out*Originally posted by genius***In all honesty, your answer is sufficient. Take the Fibonacci sequence, for instance, which takes the form 1,1,2,3,5,8...**

Normally you would write it as a recurrence relation,

f[n]=f[n-2]+f[n-1].

However, we can also write it as

f[n]=((A^n)-(1-A)^n)/sqrt(5)

where A=(1+sqrt(5))/2. Writing it in any other form would be unnecessary.

a(n) is the ratio used as the average distance for n > or = to 3 - 21 Nov '07 10:14 / 1 edit

There's a standard way of solving these sort of problems ('difference equations'. It's actually very similar to the method of solving ordinary differential equations (if you've come across those).*Originally posted by joe shmo***thanks for that, I was just going to ask about the Fibonacci sequence and I had a hunch that the golden ratio could be used to figure out**

a(n) is the ratio used as the average distance for n > or = to 3

For the Fibonnaci series: a(n+2) = a(n + 1) + a(n)

Look for a solution in the form a(n) = x^n.

x^(n + 2) = x^(n + 1) + x^n

=> x^n[x^2 - x - 1] = 0

This gives you a quadratic equation. Using the usual quadratic formula:

x = (1 +- sqrt(5))/2

But the problem is linear - if you have two solutions you can add them together and you still have a solution. So the most general solution is:

a(n) = A[(1 + sqrt(5))/2]^n + B[(1 - sqrt(5))/2]^n

To find A and B you look at the first two terms. You know a(0) = 1 and a(1) = 1.

=> A + B = 1

and (A - B)sqrt(5) = 1

Solve for A and B and you've got the answer.