Is it tedious?

Algebra Level 3

$\large x=\frac { 4 }{ \left( \sqrt { 5 } +1 \right) \left( \sqrt [ 4 ]{ 5 } +1 \right) \left( \sqrt [ 8 ]{ 5 } +1 \right) \left( \sqrt [ 16 ]{ 5 } +1 \right) }$

Find $\left( x+1 \right) ^{ 16 }$ .

The answer is 5.

2 solutions

A Former Brilliant Member
Apr 11, 2016

I will use exponent as a fraction rather than the nth root symbol because it is easier to type :P

Multiplying the numerator and denominator by all of the denominator's conjugate we get $x=\dfrac{4(5^{1/2}-1)(5^{1/4}-1){(5^{1/8}-1)} (5^{1/16}-1)}{4(5^{1/2}-1)(5^{1/4}-1)(5^{1/8}-1)}=5^{1/16}-1$

Therefore $(x+1)^{16}=(5^{1/16}-1+1)^{16}=\boxed5$ .

Rather a generalised result:

$\large\dfrac{a-1}{\displaystyle\prod_{r=1}^{n}(\sqrt[\large{2^r}]{a}+1)}=(\sqrt[2^n]{a}-1)$

And therefore $((\sqrt[2^n]{a}-1)+1)^{2^n}$ $=a$

Anyways $+1$ ..

Rishabh Jain - 5 years, 2 months ago

Nice generalization!

A Former Brilliant Member - 5 years, 2 months ago

Yeah!!Reminds me of this .Hey Shvatejas,+1 :)

Rohit Udaiwal - 5 years, 2 months ago

Nice problem, @rohit udaiwal , and nice solution, @Svatejas Shivakumar

Swapnil Das - 5 years, 2 months ago

Very nice solution. ♥

Atanu Ghosh - 5 years, 2 months ago

Abdur Rehman Zahid
Apr 11, 2016

$\begin{aligned} x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}&=\dfrac{4(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)(\sqrt[16]{5}-1)}\end{aligned}$ Repeatedly applying the Diffrence of Squares identity $(a+b)(a-b)=a^2-b^2$ to simplify the denominator,we get: $x=\frac{4(\sqrt[16]{5}-1)}{4}=\sqrt[16]{5}-1\\ \implies (x+1)^{16}=\boxed{5}$

Is it tedious?

The answer is 5.

2 solutions

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