# If x = 4/{(5^1/2 + 1)(5^1/4 + 1)(5^1/8 + 1)(5^1/16 + 1)} then evaluate (x+1)^48.This is one of the questions i got when i was preparing for NTSE. Plz solve it using tricks or simply using your mind...

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You should convert the rational powers into radicals such that:

`5^(1/2) = sqrt 5`

`5^(1/4) = (5^(1/2))^(1/2) = (sqrt 5)^(1/2) = sqrt(sqrt 5)`

`5^(1/8) = (5^(1/4))^(1/2) = sqrt(sqrt(sqrt 5))`

`5^(1/16) = (5^(1/8))^(1/2) = sqrt(sqrt(sqrt(sqrt 5)))`

You need to evaluate `(x+1)^48` , hence, you need to evaluate x such that:

`x = 4/((sqrt 5 + 1)(sqrt(sqrt 5) + 1)(sqrt(sqrt(sqrt 5)) + 1)(sqrt(sqrt(sqrt(sqrt 5))) + 1))`

You need to multiply the fraction by the following conjugates:

`x = (4*(sqrt 5- 1)(sqrt(sqrt 5)- 1)(sqrt(sqrt(sqrt 5))- 1)(sqrt(sqrt(sqrt(sqrt 5)))- 1)))*1/(((sqrt 5 + 1)(sqrt(sqrt 5) + 1)(sqrt(sqrt(sqrt 5)) + 1)(sqrt(sqrt(sqrt(sqrt 5))) + 1)(sqrt 5 - 1)(sqrt(sqrt 5) - 1)(sqrt(sqrt(sqrt 5)) - 1)(sqrt(sqrt(sqrt(sqrt 5))) - 1)`

You need to convert the following products in difference of squares such that:

`(sqrt 5 + 1)(sqrt 5- 1) = 5 - 1 = 4`

`(sqrt(sqrt 5) + 1)(sqrt(sqrt 5)- 1) = sqrt 5 - 1`

`(sqrt(sqrt(sqrt 5)) + 1)(sqrt(sqrt(sqrt 5)) - 1) = sqrt(sqrt 5) - 1`

`(sqrt(sqrt(sqrt(sqrt 5))) + 1)(sqrt(sqrt(sqrt(sqrt 5))) - 1) = sqrt(sqrt(sqrt 5)) - 1`

`x = (4*(sqrt 5- 1)(sqrt(sqrt 5)- 1)(sqrt(sqrt(sqrt 5))- 1)(sqrt(sqrt(sqrt(sqrt 5)))- 1)))* 1/(4(sqrt 5 - 1)(sqrtsqrt 5 - 1)(sqrt sqrt sqrt 5 - 1))`

Reducing duplicate factors yields:

`x = (sqrt(sqrt(sqrt(sqrt 5)))- 1)`

You need to add 1 both sides such that:

`x + 1 = sqrt(sqrt(sqrt(sqrt 5)))- 1 + 1`

Reducing like terms yields:

`x + 1 = sqrt(sqrt(sqrt(sqrt 5))) = 5^(1/16)`

You need to evaluate `(x+1)^48` such that:

`(x+1)^48 = (5^(1/16))^48 => (x+1)^48 = (5^(48/16)) = 5^3 = 125`

**Hence, evaluating `(x+1)^48` under the given conditions, yields `(x+1)^48 = 125` .**