Exponent Result
Is there a result proving/stating that:
2^n > 2^(n-1) - 2^(n-2) - ... - 2 - 1 for n > 0
2^n > 2^(n-1) - 2^(n-2) - ... - 2 - 1 for n > 0
Original post by md100
Is there a result proving/stating that:
2^n > 2^(n-1) - 2^(n-2) - ... - 2 - 1 for n > 0
2^n > 2^(n-1) - 2^(n-2) - ... - 2 - 1 for n > 0
Do you think negative powers on the right side?
If Yes
2^n > 2^(n-1) > 2^(n-1)-2^(n-2)- ...
Original post by md100
Is there a result proving/stating that:
2^n > 2^(n-1) - 2^(n-2) - ... - 2 - 1 for n > 0
2^n > 2^(n-1) - 2^(n-2) - ... - 2 - 1 for n > 0
Obvously 2^n > 2^(n-1) and your right hand side is less than 2^(n-1) so your inequality is true.
Did you mean 2^n > 2^(n-1)+2^(n-2).....+2+1 ?
This is also true.
16>8+4+2+1=15
32>16+8+4+2+1=31
You could use the formula for a geomtric series.
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