（1/2)^(1/3) * (1/4)^(1/9)* (1/8)^(1/27)*(1/16)^(1/81).=

（1/2)^(1/3) * (1/4)^(1/9)* (1/8)^(1/27)*(1/16)^(1/81).=

I don't know if you're wrong. But what I understand is that every group is 1 / 2 N times 1 / 3 n. then the final result is 1 / 2 ^ (1 + 2 + 3 + 4 +...) +n)×1/3^(1+2+3+4+…… +n)= 1/6^(n(n+1) /2)
But according to your question, it seems that it's not this. It's 1 / 2 to the 1 / 3 power. It's relatively difficult to calculate this. But it's also relatively simple. We think ha, it's just that the operation is not easy to write. But it's the same as the one above. It's that the power of every term is n (1 / 3) ^ n, And the base numbers are all 1 / 2. So it's 1 / 2 ^ [1 / 3 + 2 / 9 + 3 / 27 + 4 / 81 + n (1 / 3) ^ n] and then it can be solved by using the dislocation subtraction method. Because it's complicated, you can solve it by yourself. It's very simple. Finally, it's easy

（2/3+2/9+2/27+2/81）-（1/2+1/4+1/8+1/16）

（2/3+2/9+2/27+2/81）-（1/2+1/4+1/8+1/16）
=(54/81+18/81+6/81+2/81)-(8/16+4/16+2/16+1/16)
=80/81-15/16
=(1-1/81)-(1-1/16)
=1/16-1/81
=65/1296

Fill in 1 / 3, 2 / 9, 4 / 27, 8 / 81, (), () according to the rules

16/243 32/729
The general term is 2 ^ n / 3 ^ n

(- 81) divided by 2 and 1 / 4-4 / 9 divided by (- 16)

(- 81) divided by 2 and 1 / 4-4 / 9 divided by (- 16)
=-81*4/9-4/9*1/（-16）
=-36+1/36
=-35 and 35 / 36