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Geometric Sequences Question.

A sequence of numbers U1, U2, ... , Un, ... is given by the formula Un = 3((2/3^n)) -1 where n is a positive integer.

a) Find the values of of U1, U2 and U3.

b) Show that the sum to fifteen terms = -9.014 to 4 significant figures.

c) Prove that Un+1 = 2((2/3^n)) -1

This question is in the Edexcel AS and A Level Modular Mathematics C2 (pearson company).

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Original post by Paranoid_Glitch
A sequence of numbers U1, U2, ... , Un, ... is given by the formula Un = 3((2/3^n)) -1 where n is a positive integer.

a) Find the values of of U1, U2 and U3.

b) Show that the sum to fifteen terms = -9.014 to 4 significant figures.

c) Prove that Un+1 = 2((2/3^n)) -1

This question is in the Edexcel AS and A Level Modular Mathematics C2 (pearson company).


What have you tried so far? Which part did you get up to?
Original post by SeanFM
What have you tried so far? Which part did you get up to?
done with a). i realised that 2/3 is the common ratio but i'm still having problems trying to prove the sum. have not gotten to c). yet
Original post by Paranoid_Glitch
done with a). i realised that 2/3 is the common ratio but i'm still having problems trying to prove the sum. have not gotten to c). yet


You're looking for the sum of the first 15 terms, so the sum of u1, u2 ...and u15).

So how would you write that as a summation, and go on to solve it?

Also, is it 3∗(23)n−13 * (\frac{2}{3})^n -1?
Original post by SeanFM
You're looking for the sum of the first 15 terms, so the sum of u1, u2 ...and u15).

So how would you write that as a summation, and go on to solve it?

Also, is it 3∗(23)n−13 * (\frac{2}{3})^n -1?
Yeah it is. i tried using a(r^n-1)/(r-1) but i cannot seem to get it. i substituted r for 2/3, and a=1. could it be a that I'm substituting in wrong
Original post by Paranoid_Glitch
Yeah it is. i tried using a(r^n-1)/(r-1) but i cannot seem to get it. i substituted r for 2/3, and a=1. could it be a that I'm substituting in wrong


As it is, it is not a geometric sequence. U1 = 1, U2 = -1/3, U3 = -7/9 etc. There's no common ratio there.

You're looking for the sum of the first 15 terms of . If you were
to express that as a sum of n terms, what would that look like? (Not a trick question).
ar^n-3 +ar^n-2 +ar^n-1?
Original post by SeanFM
As it is, it is not a geometric sequence. U1 = 1, U2 = -1/3, U3 = -7/9 etc. There's no common ratio there.

You're looking for the sum of the first 15 terms of . If you were
to express that as a sum of n terms, what would that look like? (Not a trick question).
ar^n-3 +ar^n-2 +ar^n-1?
Original post by Paranoid_Glitch
ar^n-3 +ar^n-2 +ar^n-1?


Not quite :redface: I may have been a bit confusing.

∑i=1n(3∗(23)i−1)\displaystyle\sum_{i=1}^n (3*(\frac{2}{3})^i - 1) was what I was looking for. Do you see what I meant?

If so, is there any way of splitting this up so that you have a geometric sequence + something else?
(edited 8 years ago)
Original post by SeanFM
Not quite :redface: I may have been a bit confusing.

∑i=1n(3∗(23)n−1)\displaystyle\sum_{i=1}^n (3*(\frac{2}{3})^n - 1) was what I was looking for. Do you see what I meant?

If so, is there any way of splitting this up so that you have a geometric sequence + something else?


Gosh i am so confused sorry :frown: would it be 3*(2/3)^n+1 ;(
Original post by Paranoid_Glitch
Gosh i am so confused sorry :frown: would it be 3*(2/3)^n+1 ;(


Sorry, I'm half asleep - I may have been confusing you.



That's the sum of the first 15 terms, which is what the question is asking for (but not something that you can solve just yet, but we'll get onto that once you understand this bit).

The i=1 means that the first i value you put in the thing next to the sum is i=1. Then you put in i=2 and add that to the output that you get from i=1, and then add that to what you get for putting i=3 in... all the way to n (which is what the bit on the top means).

So can you see that that's just finding u1, u2,... u15 and adding them all together?
(edited 8 years ago)
Original post by SeanFM
Sorry, I'm half asleep - I may have been confusing you.



That's the sum of the first 15 terms, which is what the question is asking for (but not something that you can solve just yet, but we'll get onto that once you understand this bit).

The i=1 means that the first i value you put in the thing next to the sum is i=1. Then you put in i=2 and add that to the output that you get from i=1, and then add that to what you get for putting i=3 in... all the way to n (which is what the bit on the top means).

So can you see that that's just finding u1, u2,... u15 and adding them all together?

Yeah i get it, so u15 = u15+u14 +u13+ u12+....+u1.
Original post by Paranoid_Glitch
Yeah i get it, so u15 = u15+u14 +u13+ u12+....+u1.


Great :smile:

Can you see a way to split that into two seperate sums? (Hint: One is a geometric sum)
Original post by SeanFM
Great :smile:

Can you see a way to split that into two seperate sums? (Hint: One is a geometric sum)
sorry i just cannot figure this out. could it be a(1-r^n) and a(1-r^n)/1-r? where a=2 and r=2/3. i really do not know sorry :frown:
Original post by Paranoid_Glitch
sorry i just cannot figure this out. could it be a(1-r^n) and a(1-r^n)/1-r? where a=2 and r=2/3. i really do not know sorry :frown:


Those come in if you have a geometric sequence.


You said that you understand how we got to




Let me know if you want me to explain that again.


We can't apply the geometric formulae that you've mentioned because that is not a geometric sequence. There is no common ratio between u1 and u2.

But we can notice however that for each term, you are taking away 1. So if you are summing the 15 terms, then you have for each term, something and then -1. So that would give you (sum of somethings) -15 in total.

You can see that by splitting the sum up into two.

In other words,





is the same as
∑i=1153∗(23)i\sum_{i=1}^{15} 3*(\frac{2}{3})^i + ∑i=115−1\sum_{i=1}^{15}{-1} , and the sum of 15 lots of -1 is -15, and the first term looks more familiar.

Let me know if any of that needs more explaining. :smile:
(edited 8 years ago)
Original post by SeanFM
Those come in if you have a geometric sequence.


You said that you understand how we got to




Let me know if you want me to explain that again.


We can't apply the geometric formulae that you've mentioned because that is not a geometric sequence. There is no common ratio between u1 and u2.

But we can notice however that for each term, you are taking away 1. So if you are summing the 15 terms, then you have for each term, something and then -1. So that would give you (sum of somethings) -15 in total.

You can see that by splitting the sum up into two.

In other words,





is the same as
∑i=1153∗(23)i\sum_{i=1}^{15} 3*(\frac{2}{3})^i + ∑i=115−1\sum_{i=1}^{15}{-1} , and the sum of 15 lots of -1 is -15, and the first term looks more familiar.

Let me know if any of that needs more explaining. :smile:
oh i think I'm understanding it. so we get the sum of the first fifteen terms then subtract -1 for each term, and because it is 15 terms its -15 because for Un we are subtracting 1 from each term, is that what you mean.
Reply 16
Avatar for Zacken
Zacken
22
Original post by Paranoid_Glitch
oh i think I'm understanding it. so we get the sum of the first fifteen terms then subtract -1 for each term, and because it is 15 terms its -15 because for Un we are subtracting 1 from each term, is that what you mean.


Pretty much! It's a more general rule of if you have

∑(a+b)=∑a+∑b\sum (a+b) = \sum a + \sum b

By the way, I'll be moving this to the maths forum for you!
Original post by Paranoid_Glitch
oh i think I'm understanding it. so we get the sum of the first fifteen terms then subtract -1 for each term, and because it is 15 terms its -15 because for Un we are subtracting 1 from each term, is that what you mean.


That's one way of explaining it yes, but the important thing is to see that you can do it by splitting up the sum.

Eg the sum of ax^2 + bx + c is a* the sum of x^2 + b*sum of x + sum of c. (where the sum of n lots of c is nc).

So you've got the -15 out of the sum, how can you calculate the sum that's left over?
Original post by Zacken
Pretty much! It's a more general rule of if you have

∑(a+b)=∑a+∑b\sum (a+b) = \sum a + \sum b

By the way, I'll be moving this to the maths forum for you!
oh okay thanks
Original post by SeanFM
That's one way of explaining it yes, but the important thing is to see that you can do it by splitting up the sum.

Eg the sum of ax^2 + bx + c is a* the sum of x^2 + b*sum of x + sum of c. (where the sum of n lots of c is nc).

So you've got the -15 out of the sum, how can you calculate the sum that's left over?


I got the answer using: 3*((2/3(2/3^15 - 1))/(2/3-1) though i do not understand why is it because U1 = 3*(2/3^1)?
(edited 8 years ago)

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