Show that the terms of the series are in geometric progression.

[alexapprend]

Hello, here is the question from the textbook:

The sum of [math]n[/math] terms of a certain series is given by [imath](\frac{3}{2})^2 -1\\[/imath]
a) find the firt 3 terms in the series (I have done this, but substituting 1, 2, 3 etc and the finding the difference between the sum of 1 and sum of 2)
------[math]\frac{1}{2}, \frac{3}{4}, \frac{9}{8}\\[/math]b) Show that the terms of the series are in geometric progression

My first instinct here is to just find the difference between terms and show that it is consistent (or even just it back in to show that is matches the formula for a geometric progression) but the worked solutions show that it wants you to get it back into the form of a geometric series equation in a different way. Formula we're aiming for:

This is the worked solution:

I can understand the first line shows the same things expanded/factorised but I don't understand where they that line from- it looks like the sum for when n=1 (because then the second term (3/2)^(n-1) would be the power of zero and therefore be 1) but how does that work? You just get to use the sum equation? Is that a known thing that I missed. I also don't understand how the first line can become the second line (I understand taking a half out but I don't understand 3/2 disappearing and 3/2^n-1 staying the same.

Sorry for the dodgy writing of equations, I'm still working out latex. For context: this is IB HL maths- I am self teaching from the textbook. I think this is the correct forum.

 

[Dr.Peterson]

Hello, here is the question from the textbook:

The sum of [imath]n[/imath] terms of a certain series is given by [imath](\frac{3}{2})^2 -1\\[/imath]
a) find the firt 3 terms in the series (I have done this, but substituting 1, 2, 3 etc and the finding the difference between the sum of 1 and sum of 2)
------[imath]\frac{1}{2}, \frac{3}{4}, \frac{9}{8}\\[/imath]
b) Show that the terms of the series are in geometric progression

I think you meant [imath](\frac{3}{2})^n -1[/imath], right?

I can understand the first line shows the same things expanded/factorised but I don't understand where they that line from- it looks like the sum for when n=1 (because then the second term (3/2)^(n-1) would be the power of zero and therefore be 1) but how does that work? You just get to use the sum equation? Is that a known thing that I missed. I also don't understand how the first line can become the second line (I understand taking a half out but I don't understand 3/2 disappearing and 3/2^n-1 staying the same.

The first thing they do, which I think is what you did for specific terms, is to subtract the sum of n-1 terms from the sum of n terms to find the nth term.

Then they factor out [imath]\left(\frac{3}{2}\right)^{n-1}[/imath] from the two terms. Then, to get to the second line, they just subtract 3/2 - 1 = 1/2.

Which part do you not understand?

 

[alexapprend]

Yes, sorry I meant to the power of n for the sum. I can follow what you're saying and I now see I was pretty dumb to miss that they simplfied 3/2-1. It took me a while to see what you're saying about the first step- I did not think about it algebraically, I just went straight into substituting the numbers 1, 2, 3 etc. but from what you said (and sitting with it for a while) I realised that my answers on the first step would be expressed in that way and the -1 disappears because it's -1-(-1). Anyway, thank you so much for your help- I'm struggling through this book one question at a time. Proofs are one of things I struggle with most and I truely wish the worked solutions wouldn't skip so many steps- my brain can't yet fill in the gaps but I'm working on it.

 

[Dr.Peterson]

Yes, sorry I meant to the power of n for the sum. I can follow what you're saying and I now see I was pretty dumb to miss that they simplfied 3/2-1. It took me a while to see what you're saying about the first step- I did not think about it algebraically, I just went straight into substituting the numbers 1, 2, 3 etc. but from what you said (and sitting with it for a while) I realised that my answers on the first step would be expressed in that way and the -1 disappears because it's -1-(-1). Anyway, thank you so much for your help- I'm struggling through this book one question at a time. Proofs are one of things I struggle with most and I truely wish the worked solutions wouldn't skip so many steps- my brain can't yet fill in the gaps but I'm working on it.

The difference between what you did and what they did is that you only showed that the first three terms are in geometric sequence, while they prove that all terms are. And starting as you did with specific numbers is an excellent strategy for discovering a proof, or in general for understanding how any kind of math works. Concrete examples are easier to follow than abstract stuff.

And one thing we can help with is filling in gaps like this for you. Make a start, show us what you've done, and we'll gladly help you with the next bit.