SOLUTION: List the first six terms of the sequence defined by bn= (2*4*6 …2n)/2^n.

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Question 1091272: List the first six terms of the sequence defined by bn= (2*4*6 …2n)/2^n.

Answer by ikleyn(52788) About Me  (Show Source):
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List the first six terms of the sequence defined by bn= (2*4*6 …2n)/2^n.
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1-st term, n= 1:  b%5B1%5D = 2%2F2%5E1 = 2%2F2 = 1.

2-nd term, n= 2:  b%5B2%5D = %282%2A4%29%2F2%5E2%7D%7D+=+%7B%7B%7B8%2F4 = 2.

3-nd term, n= 3:  b%5B3%5D = %282%2A4%2A6%29%2F2%5E3%7D%7D+=+%7B%7B%7B48%2F8 = 6.

4-th term, n= 4:  b%5B4%5D = %282%2A4%2A6%2A8%29%2F2%5E4%7D%7D+=+%7B%7B%7B384%2F16 = 24.


Can you complete the assignment from this point on your own ?