Readymade Radicals

Algebra Level 4

$\large \sqrt[3]{1\sqrt[3]{16\sqrt[3]{256\cdots}}}= \ ? \$

Clarification : $1,16,256,\ldots$ follows a geometric progression .

3 2 8 1 4

3 solutions

Akshat Sharda
Feb 9, 2016

$\begin{aligned} & = \sqrt[3]{1\sqrt[3]{16\sqrt[3] {256\cdots}}} \\ & = (1(16(256(\ldots)^\frac{1}{3})^\frac{1}{3})^\frac{1}{3})^\frac{1}{3} \\ & = 16^\frac{1}{3^2}\cdot 16^\frac{2}{3^3}\cdot 16^\frac{3}{3^4} \ldots \\ & = 16^{\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\ldots } \\ & = 16^{\frac{1}{4}}=\boxed{2} \end{aligned}$

$\frac {1}{3^2}$ + $\frac {2}{3^3}$ + ... = $\frac {1}{4}$ Can you please explain this. haha

Tim Kristian Llanto - 5 years, 4 months ago

$\begin{aligned} S &= \frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\ldots \\ \frac{S}{3} & = \quad \quad \frac{1}{3^3}+\frac{2}{3^4}+\ldots \\ S-\frac{S}{3}=\frac{2S}{3} & = \frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\ldots \\ S & =\frac{3}{2} \times \frac{\frac{1}{3^2}}{1-\frac{1}{3}} = \frac{1}{4} \end{aligned}$

Akshat Sharda - 5 years, 4 months ago

Nemesis Coming
Feb 4, 2019

Question= 2^(4/9+8/27+16/81+......)

Now here taking common 4/9 in powers

We get by taking only power terms 4/9(1+2/3+3/9+4/9+5/81...)

By solving A=1+2/3+3/9+4/27+5/81+6/243..... 3A=3+2+3/3+4/9+5/27+6/81+...

By subtracting same demonstrator terms

2A=3+1+1/3+1/9+1/27+1/81.... By gp 2A=3+1/(1-1/3) 2A=9/2 A=9/4

By putting it's value 4/9(9/4)=1 So answer is 2

Aman Rckstar
Feb 16, 2016

Simple , just solve it by taking all numbers in powers of 2 , Now it formed a AGP as 2^( 4/9 + 8/27 + 12/81 ...) Sove to get value of AGP as 1 and as 2^(1) = 2

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