+0

helppppp

+1
321
3

What real number is equal to the expression $$2 + \frac{4}{1 + \frac{4}{2 + \frac{4}{1 + \cdots}}}$$, where the 1s and the 2s alternate?

May 8, 2018

#1
+5

Let  a   =  2  +   4

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1 +   4

____

........

So we have

a =   2  +       4

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1  + 4

__

a           simplify

a =   2      +    4a

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(a  + 4)

a   =  (2a + 8 + 4a)

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(a + 4)

a( a + 4)  =   6a + 8

a^2 + 4a   = 6a  + 8

a^2 - 2a  - 8   = 0

(a - 4) (a + 2)    = 0

Setting both factors to 0  and solving for   "a"  we  get   that  a  = 4  or a  = -2

We can reject -2 because  a  is positive

So......the real  number is   4   May 8, 2018
#2
+2

Thank you, CPhill! ant101  May 8, 2018
#3
+3

Solution:

So, we have $$2+\frac{4}{1+\frac{4}{2+\frac{4}{1+\cdots}}}$$, which at first looks rather scary. But we notice a pattern! Because the nested fractions descend forever, the fraction essentially contains itself! So let $$a$$  equal this fraction:

$$a = 2+\frac{4}{1+\frac{4}{2+\frac{4}{1+\cdots}}}$$

Thus,

$$a = 2+\frac{4}{1+\frac{4}{a}}$$

So now, we simply need to solve for a.

$$a = 2+\frac{4}{1+\frac{4}{a}} = 2 + \frac{4}{\frac{4+a}{a}}= 2+\frac{4a}{4+a}=\frac{8+2a+4a}{4+a}=\frac{8+6a}{4+a}$$

Getting rid of the fraction by multiplying y $$4+a$$, we get

$$4a + a^2 = 8 + 6a \Longrightarrow a^2 - 2a - 8 = 0$$

So now we factor...

$$a^2 - 4a + 2a - 8 = 0$$

$$a(a-4)+2(a-4) = 0$$

$$(a+2)(a-4) = 0$$

So $$a$$ can be $$-2$$ or $$4$$. Since we know, by looking at the fraction, that $$a$$  must be positive, $$a$$, and the value of the fraction, must be $$\boxed{4}$$.

Tell me if you have any questions,  May 8, 2018