SOLUTION: The infinite geometric sequence which begins: (2^300),(2^298),(2^296),(2^294)... contains only ONE odd integer. This odd integer is the nth term of the sequence. What is the

Algebra ->  Sequences-and-series -> SOLUTION: The infinite geometric sequence which begins: (2^300),(2^298),(2^296),(2^294)... contains only ONE odd integer. This odd integer is the nth term of the sequence. What is the       Log On


   



Question 73595: The infinite geometric sequence which begins:
(2^300),(2^298),(2^296),(2^294)...
contains only ONE odd integer. This odd integer is the nth term of the sequence. What is the value of n?
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I figured out the equation for the sequence, but don't know how to find out the odd integer.
equation: (2^302-2n)

Answer by joyofmath(189) About Me  (Show Source):
You can put this solution on YOUR website!
2%5Ex is always going to be even except when x+=+0 in which case 2%5Ex+=+2%5E0+=+1.
So, when is 302-2n+=+0? That'd be when 302+=+2n or n+=+151.