To find f(n+1) - f(n-1) in-terms of f(n)

[gmattesttaker2]

Hello,

Can you please tell me how to solve this:

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)?


A) 1/2 (f(n))

B) sq. root ( f(n) )

C) f(n)

D) 2 f(n)

E) ( f(n) )^2

OA: [spoiler]f(n)[/spoiler]


f(n) = (1 + sq. root 5)^n/2^n = ( (1 + sq. root 5)/2 )^n

f(n + 1) = ( (1 + sq. root 5)/2 )^(n + 1)
f(n - 1) = ( (1 + sq. root 5)/2 )^(n - 1)

f(n + 1) - f(n - 1) = ( (1 + sq. root 5)/2 )^(n + 1) - ( (1 + sq. root 5)/2 )^(n - 1)

I was not able to solve from here onwards. The official explanation mentions about some exponent rule of factoring out the smaller power but I am not very familiar with that. Can you please assist? Thanks a lot for your help.

Best Regards,
Sri

[GMATGuruNY]

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)?

A) 1/2 (f(n))

B) sq. root ( f(n) )

C) f(n)

D) 2 f(n)

E) ( f(n) )^2

The constraint that n>2 is irrelevant, since there is no value of n that will yield a nonreal value for f(n) = (1+√5)^n/2^n.
Thus, we can plug in ANY VALUE for n.

Let n=1.
Then:
f(n+1) = f(2) = (1+√5)²/2² = (1 + 2√5 + 5)/4 = (6 + 2√5)/4 = (3 + √5)/2.
f(n-1) = f(0) = (1+√5)�/2� = 1/1 = 1.
Thus:
f(n+1) - f(n-1) = [ (3 + √5)/2 ] - 1 = [ (3 + √5)/2 ] - (2/2) = (1 + √5)/2. This is our target.

Now plug n=1 into the answers to see which yields our target of (1 + √5)/2.
Since all of the answer choices are in terms of f(n) -- the value in answer choice C -- start with C:
f(n) = f(1) = (1+√5)¹/2¹ = (1 + √5)/2.
Success!

The correct answer is C.

[feedrom]

Hi Mitch,

The question said that: If n is a positive integer greater than 2, so it's necessary to n>2?

Thank you,

[GMATGuruNY]

Hi Mitch,

The question said that: If n is a positive integer greater than 2, so it's necessary to n>2?

Thank you,

I've amended my post above to include the following:

The constraint that n>2 is irrelevant, since there is no value of n that will yield a nonreal value for f(n) = (1+√5)^n/2^n.
Thus, we can plug in ANY VALUE for n.