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Mathematics 18 Online
OpenStudy (anonymous):

what is it series for the following x=1 , 2 , 4 , 64 --- and so on

OpenStudy (amistre64):

what is 'what'?

OpenStudy (anonymous):

x(sub n) = 2^(n-1) is the best that I have. It isn't right for the set you wrote, but are you sure it isn't 1,2,4,...,64?

OpenStudy (amistre64):

the sequence might be: A{n} = 1 + A{n-1}^ .... thats where my confusion is too lol

OpenStudy (amistre64):

2^2 = 4 4^3 = 64 5^4 = ... 6^5 = ...

OpenStudy (amistre64):

only thing is... 1^2 = 1 :)

OpenStudy (anonymous):

1^1 2^2 4^3 8^4?

OpenStudy (amistre64):

ack 64^4 = .... ...^5 = .... ... ^6 maybe?

OpenStudy (anonymous):

which is A{n}=(2^(n-1))^n

OpenStudy (anonymous):

OR A{n}=2^(n^2-n)

OpenStudy (anonymous):

Final answer

OpenStudy (amistre64):

i think thats the best we can do without knowing what the problem is asking for to begin with :)

OpenStudy (anonymous):

2^0, 2^1, 2^2, 2^6 best I can do is 2^16 (16 is twice the sum of the previous 2 exponents) the question lacks definition.

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