Calculating a sequence

[u/Beehive-40]

Hi all! I'm trying to figure something out. I want to learn how to calculate something, but don't know the terminology to figure out how to find an answer. What I want to calculate is as follows:

(x+3) * 1.1 = x

X begins at 8, and I want to repeat/graph this out to have in a table for reference.

So basically, I want to take 8, add 3, multiply by 1.1, then do it again any number of times. I can just keep inputting +3*1.1 in a calculator, but there must be a better method.

Any help would be appreciated :)

[u/ProctologyExplosion]

Perfect task for a spreadsheet.

[u/Mu_Lambda_Theta]

Let x be the starting value.

x_1 = (x + 3)*1.1 = 1.1x + 3.3

x_2 = (x_1 + 3)*1.1 = ((x + 3)*1.1 + 3)*1.1 = 1.21x + 6.93

x_3 = (x_2 + 3)*1.1 = ((x_1 + 3)*1.1 + 3)*1.1 = (((x + 3)*1.1 + 3)*1.1 + 3)*1.1 = 1.331x + 10.923

The terms are always of the form x*1.1^n + (some number)

That number is always increased by 3 and multiplied by 1.1, so it's effectively the sequence you'd get if the starting value were 0.

3*1.1 ==> (3*1.1 + 3)*1.1 = 3*1.1^2 + 3*1.1 ==> (3*1.1^2 + 3*1.1 + 3)*1.1 = 3*1.1^3 + 3*1.1^2 + 3*1.1

You can probably see the pattern. It's a bunch of powers of 1.1 multiplied by 3. How much? Exactly n many; x_1 would be 3*1.1, x_2 would be 3*(1.1^2 + 1.1), etc.

We know that k^n + k^(n-1) + k^(n-2) + ... + k^2 + k^1 is equal to k*(k^n-1)/(k-1), which is a variant of a finite geometric series (that starts a bit later). In our case, k = 1.1.

So, we get that the extra number is 3 * 1.1*(1.1^n - 1)/(1.1-1) = 33*(1.1^n - 1)

The nth term of the sequence is equal to:

1.1^n * x + 33*(1.1^n - 1)

Because you said you started with x=8:

41 * 1.1^n - 33

(Edited to mark as spoilers, just in case)

[u/MontechApps]

That's called a recurrence relation. For a table, a spreadsheet is the easiest way. Put 8 in A1, type =(A1+3)*1.1 in A2, then drag it down as far as you want. If you want a formula that jumps straight to any step, it works out to 41 × 1.1ⁿ − 33, where n is the number of repeats. So n = 0 gives 8, n = 1 gives 12.1, n = 2 gives 16.61. The 33 comes from solving x = (x+3) × 1.1, which gives x = −33. That's the value the sequence would stay at if you started on it. Your starting number's distance from it (8 + 33 = 41) then just grows by 1.1 each step. Searching for arithmetico-geometric sequences will turn up more on this.

[u/seanv507]

so write out the first few terms in terms of x_0 (8)

I make the sequence x_n = x_0 * (f)^n +c*f * \sum_{i=0}^{n-1} f^i

where x_0 is 8 f=1.1 and c=3

the sum term is a geometric series (and has a formula for it)

[u/adrian_p_morgan]

If you're looking for terminology, try "iteration". As for practical methods, I agree with the people who suggest using a spreadsheet.