```Question 614717
<pre>
Wr'll find the explicit formula first:

Sequence 1 = {{{2/3}}}, {{{4/9}}}, {{{16/81}}}, ...

That's the sequence

a<sub>1</sub>={{{(2/3)^1}}}, a<sub>2</sub>={{{(2/3)^2}}}, a<sub>3</sub>={{{(2/3)^4}}}, ...

The exponents follow the sequence

Sequence 2:  1, 2, 4, ...

That's the sequence:

2<sup>0</sup>, 2<sup>1</sup>, 2<sup>2</sup>, ...

Those exponents follow the sequence:

Sequence 3:  0, 1, 2

which follow the sequence

1-1, 2-1, 3-1

which has nth term:

n-1

Therefore sequence 2 has nth term:

2<sup>n-1</sup>

Therefore sequence 1 has sequence:

a<sub>n</sub> = {{{(2/3)^(2^(n-1))}}}

That's the explicit formula.

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So the sequence is

{{{(2/3)^(2^0)}}}, {{{(2/3)^(2^1)}}}, {{{(2/3)^(2^2)}}}, ...

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To find a recursion formula:

Divide the (n+1)st term by the nth term:

{{{(2/3)^(2^((n+1)-1))}}}÷{{{(2/3)^(2^(n-1))}}}

{{{(2/3)^(2^(n))}}}÷{{{(2/3)^(2^(n-1))}}}

{{{(2/3)^(2^n-2^(n-1))}}}

{{{(2/3)^(2^(n-1)(2-1))}}}

{{{(2/3)^(2^(n-1)(1))}}}

{{{(2/3)^(2^(n-1))}}}

a<sub>n+1</sub> = a<sub>n</sub>·({{{2/3}}})<sup>2<sup>n-1</sup></sup>

And the complete recursion formula is:

a<sub>1</sub> = {{{2/3}}}, a<sub>n+1</sub> = a<sub>n+1</sub> = a<sub>n</sub>·({{{2/3}}})<sup>2<sup>n-1</sup></sup>

Edwin</pre>```