Urgent HL homework - Geometric Sequences

[astonky]

How can I do this:

S_n is the sum of the first n terms of a geometric sequence with first term a and common ratio r. Let P_n represent the product of these terms. Write P_n in terms of a and r.

Show that the sequence formed by taking the reciprocals of the terms is also geometric. Write R_n, the sum of the first n terms of the reciprocals, in terms of a and r, and hence show that
(S_n/R_n)ⁿ = (Pn)²

[kw0573]

S_n you should know, and is in the data booklet. P_n = a * (ar) * (ar²) * ... * a(r^n-1) = aⁿ(r° * r¹ * r^² * ... r^n-1) = aⁿ(r^{0 + 1 + 2 + ... n-1})

= aⁿ * r ^ ((n-1)(n) / 2)).
The reciprocal sequence is 1/a, 1/(ar), 1/(ar²), you need to show that this is in the form t_n=(b * s^n-1), where b is analogous to a, and s is analogous to r. 

Once you get the general term, plug it into the formula for geometric series. 

Let me know where exactly do you need help with. 
 

[astonky]

Thanks a lot for your quick response. I'll shoot a few more questions at you, if you don't mind.

 

I''m good with S_n. I get a part of P_n but I though it would be aⁿ * r^n-1 instead of aⁿ * r ^ ((n-1)(n) / 2)). Can you please explain that?

And what do you mean by showing this in the form t_n=(b* s^n-1 ), like t_n= 1/ (a * r^n-1) ?

So would R_n = (1/a ( 1- 1/ rⁿ)) / (1-r) ?

After that, I'm really confused. What do you do next?

 

[Vioh]

2 hours ago, astonky said:

I''m good with S_n. I get a part of P_n but I though it would be aⁿ * r^n-1 instead of aⁿ * r ^ ((n-1)(n) / 2)). Can you please explain that?

Note that 0 + 1 + 2 + ... n-1 is simply an arithmetic series, which means that it's equal to n(n-1)/2. Here is my full worked solution for your question. Let X represent the first sequence, and Y represent the second (which is the reciprocal of the first). Since X is a geometric sequence, so X_n=ar^n-1 by definition.

(1) Because P_n is the product of the first n terms so we have: P_n=(ar⁰)(ar¹)...(ar^n-1)=aⁿr^{0+1+...+(n-1)}. As mentioned, 0,1,...,(n-1) is an arithmetic series, so the answer must be:

P_n = aⁿr^0+1+...+n-1 = aⁿr^(n(n-1)/2)

And hence:

(P_n)² = a²ⁿr^n(n-1)

(2) Because Y is the reciprocal of the first, so Y_n=1/X_n=1/(ar^n-1)=(1/a)(1/r)^n-1. It's now clear that the Y must also be a geometric sequence, with the first term being (1/a), and the common ratio being (1/r).

(3) Substituting (1/a) for the first-term and (1/r) for the common ratio into the formula for finding the sum of a geometric series, we have: R_n=(1/a)(1-(1/rⁿ))/(1-r). Simplifying this whole expression is left as an exercise for you. But all in all, you should be able to get:

R_n = (1/a) * { r(rⁿ-1) } / { rⁿ(r-1) }

(4) As for S_n, it is straight out of the formula booklet: S_n=a(rⁿ-1)/(r-1). Hence:

S_n/R_n = {a(rⁿ-1)/(r-1)} / {(1/a)*r(rⁿ-1)/rⁿ(r-1)} = {a(rⁿ-1)*a*rⁿ(r-1)} / {r(rⁿ-1) (r-1)}

=> S_n/R_n = a²r^n-1

=> (S_n/R_n)ⁿ= (a²r^n-1)ⁿ = a²ⁿr^n(n-1)

This completes the proof.

Edited August 10, 2016 by Vioh