Question

2. What is the sum of the geometric sequence 4, 12, 36 . if there are 9 terms?

Answer 1

i know the common ratio is 3, i just dont know how to set up and solve.

Answer 2

well, you see that f(n+1) = 3 * f(n), so x = 3. I'll just call the common ratio x. The formula for the sum of the geometric series is \[\sum_{k=0}^{n} x^n = \frac{1-x^{n+1}}{1-x}\]
Multiplying that with the first term of the sequence will yield what we were looking for. Do you think you can take it from here? Would you be interested in the derivation fo this formula?

Answer 3

*of this formula

Answer 4

Yes that looks more like a sigma lesson for honnors.

Answer 5

honors*

Answer 6

Can you rephrase that sentence? I've got no clue what the hell that means. Sorry, my native language isn't English

Answer 7

Really, what is it, if you dont mid my asking?

Answer 8

For an honors class they did this lesson on sigmas(i dont know what they are)
I know that theyre is a simpler formula.

Answer 9

there*

Answer 10

It's just a way to write down a sum. Alternatively, but absolutely equivalent: Let's say we have a sum looking like

\[1 + x + x^2 + x^3 + ... + x^n \] We can see that the any summand is the previous summand multiplied by x. Now there's this formula to calculate sums like this:\[1 + x + x^2 + x^3 + ... + x^n = \frac{1-x^{n+1}}{1-x}\]
I could derivate this formula for you, if you like, or should I rather move on and show how you can find the answer to your question using it?

Answer 11

Can you show me how to find the answer?

Answer 12

Okay. So what we were looking for is

\[b + bx + bx² + ... + bx^n = b\cdot (1 + x + x² + ... + x^n)\]
As the general way of writing it down. Sorry I can't help it, I'm a nerd.
For b = 4, x=3 and n = 9. Can you maybe think of a way to calculate this using the formula I gave?

Answer 13

ohh my this looks so confusing. :P Thats okay thank you for trying to help me ill give you best answer anyway

Answer 14

Sry, if we count b as a term to, then n would be 8 of course

Answer 15

Well okay then I'm just going to finish what I started. Again, I suppose that we know:
\[1 + x + x^2 + x^3 + ... + x^n = \frac{1-x^{n+1}}{1-x}\]We want to know the value of\[b + bx + bx² + ... + bx^n = b\cdot (1 + x + x² + ... + x^n)\]
So, plugging in the formula from above into the one from below we get\[b \cdot (1+x+x²+...+x^n) = b \cdot \frac{1-x^{n+1}}{1-x}\]
Now we just have to plug the specific values of our problem in and we're done.
The starting term of our sequence is b = 4, x = 3 is the common ratio and we want to calculate a series with 9 terms altogether, so we get n = 8. We get\[ b \cdot \frac{1-x^{n+1}}{1-x} = 4 \cdot \frac{1-3^{8+1}}{1- 3} = 9841\]

Answer 16

Maybe this will help you: What we were looking for was\[4 + 4\cdot 3 + 4\cdot 3^2 + 4 \cdot 3^3 + 4 \cdot 3^4 + 4\cdot 3^5 + 4 \cdot 3^6 + 4 \cdot 3^7 + 4\cdot 3^8 \]As you can see I just took the first 9 terms of the geometric sequence and added them together. From this on I think my solution from above really shows every step