yourminsx049
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I know that 2^0+2^1 = 1 + 2 = 3 = 2^2 – 1
What does this have to do with the binary number circle?
- Thread starter yourminsx049
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I know that 2^0+2^1 = 1 + 2 = 3 = 2^2 – 1
Dr.Peterson
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What do you mean by "the binary number circle"? I tried searching for the phrase, and didn't find anything that looked relevant.
But this is the formula for a finite geometric series with r=2, and in binary it represents the fact that, for example, 11111111 = 100000000 - 1.
But this is the formula for a finite geometric series with r=2, and in binary it represents the fact that, for example, 11111111 = 100000000 - 1.
pka
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Here is the standard proof.
\(\begin{array}{*{20}{l}}{\sum\limits_{k = 0}^n {{2^k}} }& = &{1 + 2 + {2^2} + \cdots + {2^n}} \\ {2\sum\limits_{k = 0}^n {{2^k}} }& = &{2 + {2^2} + {2^3} + \cdots + {2^n} + {2^{n + 1}}} \\ \hline {\sum\limits_{k = 0}^n {{2^k}} }& = &{{2^{n + 1}} - 1} \end{array}\)
If you answer Prof Peterson's question, we can see if more is needed.
\(\begin{array}{*{20}{l}}{\sum\limits_{k = 0}^n {{2^k}} }& = &{1 + 2 + {2^2} + \cdots + {2^n}} \\ {2\sum\limits_{k = 0}^n {{2^k}} }& = &{2 + {2^2} + {2^3} + \cdots + {2^n} + {2^{n + 1}}} \\ \hline {\sum\limits_{k = 0}^n {{2^k}} }& = &{{2^{n + 1}} - 1} \end{array}\)
If you answer Prof Peterson's question, we can see if more is needed.