How Is the Sequence 2, 0, 2/27, 0, 2/125 Formulated?

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The sequence 2, 0, 2/27, 0, 2/125 can be formulated by recognizing that the odd terms follow the pattern 2/n^3, while the zeros represent every other term. A suggested representation for the non-zero terms is 1 - (-1)^n, which effectively alternates between values. Another sequence discussed, 1, 4, 1, 16, 1, 36, has its odd terms defined as f(2n + 1) = 1 and even terms as f(2n) = 4n^2. The discussion emphasizes defining sequences through systematic patterns, particularly using odd and even indices for clarity.
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Can someone help me with this sequence?

2 0 2/27 0 2/125

If I only look at the odd numbers it's: 2/n^3

But I don't know how they get that zero..
 
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Could be just every other term is 0.

A convenient way to write that would be 1 - (-1)^n in place of the 2.
 


CRGreathouse said:
Could be just every other term is 0.

A convenient way to write that would be 1 - (-1)^n in place of the 2.

Thank you very much!
 


CRGreathouse said:
Could be just every other term is 0.

A convenient way to write that would be 1 - (-1)^n in place of the 2.

Can you help me with this sequence also?

1 4 1 16 1 36

If I only look at the odd numbers it's: n^2

But I don't know how they get to '1'..
 


Alexx1 said:
Can you help me with this sequence also?

1 4 1 16 1 36

If I only look at the odd numbers it's: n^2

But I don't know how they get to '1'..

Again, just define it that way: f(2n + 1) = 1, f(2n) = 4n^2. You can use the same trick with (-1)^n if you want -- make the exponent 0 for odds and 1 for evens.
 


CRGreathouse said:
Again, just define it that way: f(2n + 1) = 1, f(2n) = 4n^2. You can use the same trick with (-1)^n if you want -- make the exponent 0 for odds and 1 for evens.

How do you define f(n)?
 


He just did. If n is odd, n= 2m+ 1 for some integer m so f(n)= 1, if n is even, n= 2m for some integer m so f(n)= f(2m)= 4(m)^2. Of course, if n= 2m then m= n/2 so you could also write this as 4(n/2)^2= 4n^2/4= n^2.
 

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