Infinite Surds

An = √(1+ An -1)                        

An = √(1+ An )

An2  = 1 + An

Quadratic formula :

a = 1   b = -1   c = -1

An = -b  ±√(b2-4ac) / 2a

An = 1  ±√(-1)2-4(1)(-1)) / 2(1)

An = 1  ±√(5) / 2

1. 1  +√(5) / 2
2. 1  -√(5) / 2 < 0
3. answer: 1  +√(5) / 2

An2  - An - 1 = 0  : As n gets very large An - An-1 = 0 ; An = An-1

2.) √(2+√2+√2+√2+…)

• As n increases, An also increases, however when n gets very large An - An  + 2

• An = √(2+ An -1)

Therefore:

• An+1 = √(2+ An )                        

An = √(2+ An -1)                        

An = √(2+ An )

An2  = 2+ An

Quadratic formula :

a = 1   b = -1   c = -2

An = -b  ±√(b2-4ac) / 2a

An = 1  ±√(-1)2-4(1)(-2)) / 2(1)

An = 1  ±√(9) / 2

        1 ± 3 / 2 = 4/2

Answer : An = 2

An2  - An - 2 = 0  : As n gets very large An - An-1 = 0 ; An = An-1

2.) √(k+√k+√k+√k+…)

        A1 = √(k+√k)

        An = √(k+ An-1)

Therefore:

        An+1 = √(k+ An)

An = √(k+ An )                        

An2  = 2+ An

An2  - An - k = 0  

Quadratic formula :

a = 1   b = -1   c = -k

An = -b  ±√(b2-4ac) / 2a

An = 1  ±√(-1)2-4(1)(-k)) / 2(1)

An = 1  ±√(1 + 4k) / 2

1+ √(1 + 4k) / 2

• In a sequence as n increases by jumping even numbers, it equals to an integer.  Sequence value of k:

• k = 2
• k = 6
• k = 12
• k = 20
• k = 30

The difference between each value is 2, 4, 6, 8, 10

• 1 + 4k needs to be a square number and it has to be odd.

So therefore if k is 2 in 1+4k then:

• When k = 2

1+4k = 1 + 4(2) = 9

• When k = 6

1+4k = 1+4(6) = 25

• 4, 9, 16, 25 , 36…

• k can not be below 0 and you can not square root a negative number so therefore.

• k :{ }