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If n is a positive integer greater than 2

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If n is a positive integer greater than 2  [#permalink]

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New post Updated on: 09 May 2018, 10:06
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If n is a positive integer greater than 2 and f(n)=\(\frac{[(1+√5)^n]}{2^n}\), what is f(n+1)−f(n−1) in terms of f(n)?

1. \(\frac{(f(n))}{2}\)
2. \(\sqrt{f(n)}\)
3. \(f(n)\)
4. \(2f(n)\)
5. \((f(n))^2\)

Would appreciate any help on how to solve this question. Thank you.

Originally posted by rhnbansal on 09 May 2018, 02:53.
Last edited by pushpitkc on 09 May 2018, 10:06, edited 1 time in total.
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If n is a positive integer greater than 2  [#permalink]

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New post 15 Aug 2018, 16:48
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Let's talk strategy here. It looks like many of the approaches in this forum focus blindly on the math, but remember: the GMAT is a critical-thinking test. For those of you studying for the GMAT, you will want to use strategies that actually minimize the amount of math that needs to be done, making it easier to manage your time. The tactics I will show you here will be useful for numerous questions, not just this one. My solution is going to walk through not just what the answer is, but how to strategically think about it. As a result, I will probably include some steps that I would normally just do in my head if it were the actual test, but I want to be as thorough as possible so you can see each step of the process. Ready? Here is the full "GMAT Jujitsu" for this question:

First, let's have a little tactical talk. We are dealing with what looks like some pretty ugly math here. (I affectionately call this "Mathugliness" in my classes.) When you see Mathugliness, you need to train your natural instincts to look for conceptual ways of thinking about the question instead of just lowering your head and plowing through the obnoxiousness. One possible way would be to plug in easy numbers. After all, the answer choices contain variables, so you can invent the value of a variable, and follow it through to the end, comparing the answer you get with the answer choices down below. Normally, this would be a good strategy for a problem like this. However, that "\(1+\sqrt{5}\)" factor in the numerator of the function isn't going to come out pretty, no matter what value for "\(n\)" you pick. You are still left juggling an enormous amount of math. So, let's start looking at a way to conceptually simplify the math instead.

The first thing you should notice is that we have an exponent distributing across the top and bottom of a fraction. Just using basic exponent rules allows you to do this:

\(f(n) = \frac{(1+\sqrt{5})^n}{2^n}=(\frac{1+\sqrt{5}}{2})^n\)

The chunk on the inside of the expression \((\frac{1+\sqrt{5}}{2})^n\) is still a little messy, so I am going to call it "\(X\)". This will help us visualize the math. Thus, \(f(n) = X^n\), where \(X=\frac{1+\sqrt{5}}{2}\).

Now, the problem is asking for "\(f(n+1) - f(n-1)\) in terms of \(f(n)\)." There is an enormous amount of leverage hidden in this prompt that we can use. First of all, when the GMAT says, "in terms of...", we know we will be using that term in the answer choices. Looking down at the answer choices not only confirms this, but it also tells us that the seemingly complicated math simplifies down to very basic relationships. As you study for the GMAT, answer choices like this should be very encouraging. You know things are going to work out.

In this case, the problem basically is asking us for \(X^{n+1} - X^{n-1}\) in terms of \(X^n\).We can factor out \(X^n\) from \(X^{n+1} - X^{n-1}\) using a strategy I call "Stealing Bases." This is a basic application of fundamental exponent rules where we factor out (or "steal") an exponential factor common to a sum (or difference) of bases.

\(X^{n+1} - X^{n-1}=(X^n)*(X^1-X^{-1})\)

Now the math is a piece of cake:
\((X^n)*(X^1-X^{-1})=f(n)*(\frac{1+\sqrt{5}}{2}-\frac{2}{1+\sqrt{5}})\)

At this stage, notice how I kept the \(f(n)\) in the equation without fully substituting that out. If you think about it, this should make a lot of sense. After all, our answer choices are "in terms of \(f(n)\)", so we want to keep that \(f(n)\) piece as is. We now can multiply both of those little fractions in the right side of the expression "by 1" to create a common denominator:

\(=f(n)*(\frac{1+\sqrt{5}}{2}*\frac{1+\sqrt{5}}{1+\sqrt{5}}-\frac{2}{1+\sqrt{5}}*\frac{2}{2})\)
\(=f(n)*(\frac{(1+\sqrt{5})^2-4}{2*(1+\sqrt{5})})\)
\(=f(n)*(\frac{1+2\sqrt{5}+5-4}{2+2\sqrt{5}})\)
\(=f(n)*(\frac{2+2\sqrt{5}}{2+2\sqrt{5}})\)
This, of course, simplifies down to \(f(n)\).

The answer is (C).

For those of you studying for the GMAT, notice all the strategies we used that helped us cut through the messiness and get our answer. These strategies are useful on a wide variety of questions, not just this one. Of course, this problem still required some math. You should never be afraid of that. But if you first look for ways to conceptually simplify problems that contain "Mathugliness", leverage the shapes of the answer choices to help you think about the question, and then only do math that gets you closer to those answer choices, you will find that many difficult GMAT questions aren't that difficult at all. Good luck, and good studying!
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Re: If n is a positive integer greater than 2  [#permalink]

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New post 09 May 2018, 05:06
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rhnbansal wrote:
If n is a positive integer greater than 2 and f(n)=[(1+√5)^n]/2^n, what is f(n+1)−f(n−1) in terms of f(n)?

1. (f(n))/2
2. √f(n)
3. f(n)
4. 2f(n)
5. (f(n))^2

Would appreciate any help on how to solve this question. Thank you.


Suppose, n = 4

f(n+1)−f(n−1)
= f(5)−f(3)
=(1+√5)^5/2^5 -(1+√5)^3]/2^3
= (1+√5)^3/2^3{(1+√5)^2/2^2 - 1}
= (1+√5)^3/2^3{(1+2√5+5-4)/2^2}
= (1+√5)^3/2^3{(2√5+2)/2^2}
= (1+√5)^3/2^3{2(√5+1)/2^2}
= (1+√5)^3/2^3{(√5+1)/2}
= (1+√5)^4/2^4
=f(4)
=f(n)

Ans: C.
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Re: If n is a positive integer greater than 2  [#permalink]

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New post 09 May 2018, 11:07
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rhnbansal wrote:
If n is a positive integer greater than 2 and f(n)=\(\frac{[(1+√5)^n]}{2^n}\), what is f(n+1)−f(n−1) in terms of f(n)?

1. \(\frac{(f(n))}{2}\)
2. \(\sqrt{f(n)}\)
3. \(f(n)\)
4. \(2f(n)\)
5. \((f(n))^2\)

Would appreciate any help on how to solve this question. Thank you.



Since f(n)=\(\frac{[(1+{\sqrt{5}})^n]}{2^n}\)

f(n+1) = \(\frac{[(1+{\sqrt{5}})^{n+1}]}{2^{n+1}} = \frac{[(1+{\sqrt{5}})^{n+1}]}{2^n * 2}\)

f(n-1) = \(\frac{[(1+{\sqrt{5}})^{n-1}]}{2^{n+(-1)}} = \frac{[(1+{\sqrt{5}})^{n-1}]}{2^n * 2^{-1}} = \frac{[2*(1+{\sqrt{5}})^{n-1}]}{2^n}\)

Now, f(n+1) − f(n−1) = \(\frac{(1+{\sqrt{5}})^{n}(1+{\sqrt{5}})^1}{2^n * 2} - \frac{2*(1+{\sqrt{5}})^{n}(1+{\sqrt{5}})^{-1}}{2^n}\) = \(\frac{[(1+√5)^n]}{2^n}(\frac{(1+{\sqrt{5}})}{2} - \frac{2}{(1+{\sqrt{5}})})\)

= \(\frac{[(1+√5)^n]}{2^n}(\frac{(1+{\sqrt{5}})^2 - 4}{2*(1+{\sqrt{5}})})\) = \(\frac{[(1+√5)^n]}{2^n}\) = f(n) - (Option C)
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Re: If n is a positive integer greater than 2  [#permalink]

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New post 27 May 2018, 07:39
f(n+1)= f(n)*(1+√5)/2

f(n-1)= f(n)*(2/1+√5)

f(n+1)-f(n-1)= f(n)*[ 1+√5/2 -2/1+√5]

Solving bracket part comes out to be 1

=f(n)

C is answer

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Re: If n is a positive integer greater than 2 &nbs [#permalink] 27 May 2018, 07:39
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