New Operation

Algebra Level 3

Let $a^{(?n)} = \sqrt[n]{a} \times \sqrt[n]{\sqrt[n]{a}} \times \sqrt[n]{ \sqrt[n]{\sqrt[n]{a}}} \times \cdots$ .

For example, $a^{(?3)} = \sqrt[3]{a} \times \sqrt[3]{\sqrt[3]{a}} \times \sqrt[3]{ \sqrt[3]{\sqrt[3]{a}}} \times \cdots$ .

If $a \in \mathbb{R^+}$ and $n \in \mathbb{Z^+}$ .

Find $3^{(?2)}$

3 2 0 9 1

1 solution

Peter Orton
Mar 24, 2015

The operation implies that

$3^{(?2)} = \sqrt{3} \times \sqrt{\sqrt{3}} \times \sqrt{ \sqrt{\sqrt{3}}} \times ......$

$3^{\frac{1}{2}} \times 3^{\frac{1}{4}} \times 3^{\frac{1}{8}} \times ........$

$3^{(1/2) + (1/4) + (1/8) + ........} = 3^1$

$=> 3 \ answer$

you should let others write solutions...but anyway.......nice question :)

Vaibhav Prasad - 6 years, 2 months ago

Please help me with my combinatorics pleaseeeeeeee!! @Vaibhav Prasad

Harsh Shrivastava - 6 years, 2 months ago

what help do you need ??/

Vaibhav Prasad - 6 years, 2 months ago

@Vaibhav Prasad – Please tell me from where should I study P and C??

Harsh Shrivastava - 6 years, 2 months ago

@Harsh Shrivastava – Simply NCERT 11th and also look up some already derived formulas on the internet

Vaibhav Prasad - 6 years, 2 months ago

@Vaibhav Prasad – There is another similar problem on brilliant and I looked up the answer from that other question

Vaibhav Prasad - 6 years, 2 months ago

Well, have you started preparing for Batch formation test .

Harsh Shrivastava - 6 years, 2 months ago

@Harsh Shrivastava – not yet i'm thinking of starting from the 1st of april

Vaibhav Prasad - 6 years, 2 months ago

@Vaibhav Prasad – Hey goku pls tell me how 1/2 +1/4 +1/8...... =1

Aman Real - 6 years, 2 months ago

@Aman Real – it is a GP

Vaibhav Prasad - 6 years, 2 months ago

@Vaibhav Prasad – How do i turn on the notification as i dont get to know who replied to me and so and so.

Aman Real - 6 years, 2 months ago

@Vaibhav Prasad – https://brilliant.org/problems/my-own-problem-highly-reputed-at-stack-exchange/

Help me with this problem.

Harsh Shrivastava - 6 years, 2 months ago

Generalising, $a^{(?n)}=\sqrt[n-1]{a}$

Omkar Kulkarni - 6 years ago