+0  
 
+1
85
3
avatar+493 

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?

ant101  May 8, 2018
 #1
avatar+87301 
+5

Let  a   =  2  +   4

                       ______

                       1 +   4

                              ____

                                 ........

                          

 

So we have

 

a =   2  +       4

                  ____

                 1  + 4

                       __

                        a           simplify

 

 

 

a =   2      +    4a

                    _______

                      (a  + 4)

 

 

a   =  (2a + 8 + 4a) 

         _____________

             (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

 

 

 

cool cool cool               

CPhill  May 8, 2018
 #2
avatar+493 
+2

Thank you, CPhill! laugh

ant101  May 8, 2018
 #3
avatar+2765 
+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,

smileysmiley

tertre  May 8, 2018

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