# Given the recursive function defined by f(n) = 3f(n-1) + 4f(n-2) for n >_ 2?

Given the recursive function defined by f(n) = 3f(n-1) + 4f(n-2) for n >_ 2

**n greater than or equal to 2**

A. 39

B. 51

C. 64

D 90

Please explain the answer for me.

Given the recursive function defined by f(n) =3f(n-1) + 4f(n-2) for n >2?

f (0) = 0

f (1) = 1

Find f(4):

A. 39

B. 51

C. 64

D. 90

### 5 Answers

- TheSicilianSageLv 79 years agoBest Answer
Given the recursive function:

f(n) =3f(n-1) + 4f(n-2) for n≥2

May be expressed explicitly as:

f(n) = c1 (-1)^n + c2 4^n

f(0) = c1 + c2 = 0

f(1) = -c1 + 4c2 = 1

or c1 = -1/5; c2 = 1/5

f(4) = -(1/5) (-1)^4 + (1/5) 4^4

.... = 51 ← Ans. B

- waidLv 43 years ago
As starwhite mentioned, this is a million, because of the fact the 2nd term is the series enlargement of exp(n). *EDIT* ok i assume it is clever that the respond must be under a million. e^n = ?{ok=0,?} n^ok / ok! which converges for all finite n. even in spite of the undeniable fact that we've a partial summation, meaning ?{ok=0,n} n^ok / ok! < e^n And as n will enhance, the discrepancy between the LHS and e^n additionally will enhance. the priority is to coach that the errors is a million/2 * e^n. *EDIT2* indexed right here are some innovations I even have to this point that i'm purely throwing accessible. Scythian stumbled on a neat relationship employing the ? applications. yet understanding you, you have some trick up your sleeve that reduces the particular applications to ordinary ones. you may write ?(n+a million,n)) / ?(n+a million) = [?{n,?} t^n exp(-t) dt] / [?{0,?} t^n exp(-t) dt] the two numerator and denominator appear like Laplace transforms, with s = a million (which does no longer impact the mixing via the way). you additionally could make a substitution to get [?{0,?} (u+n)^n exp(-u-n) du] / [?{0,?} u^n exp(-u) du] I have not got any concept the place i bypass with this, however the assumption is to coach that this decrease is a million/2 as n --> ?. *EDIT3* super discover Steiner. i'm going to ought to bypass by way of that with a magnificent enamel comb. super stuff.

- Rita the dogLv 79 years ago
f(2) = 3f(1) + 4f(0) = 3*1 + 4*0 = 3

f(3) = 3f(2) + 4f(1) = 3*3 + 4*1 = 13

f(4) = 3f(3) + 4f(2) = 3*13 + 4*3 = 51

Note: where you say n > 2 it should say n ≥ 2, otherwise you cannot do this.

- roderick_youngLv 79 years ago
I think something is missing. What is the question asking for? And does is specify what f(0) and f(1) are?

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- ?Lv 69 years ago
You haven't asked a question. You've just stated a function and some random numbers.