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vipul92

  • 3 years ago

1/3+1/6+1/12+1/24..........................infinite=?

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  1. rvgupta
    • 3 years ago
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    \[[\sqrt{(3n)^3} -2]/(9n-3)\]

  2. vipul92
    • 3 years ago
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    its wrong

  3. Zekarias
    • 3 years ago
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    \[\frac{ 1 }{ 3}+\frac{ 1 }{ 6 }+\frac{ 1 }{ 12 }+\frac{ 1 }{ 24 }+.....\] \[\frac{ 1 }{ 3}+\frac{ 1 }{ 3\times2 }+\frac{ 1 }{ 3\times4 }+\frac{ 1 }{ 3\times8 }+.....\] \[\frac{ 1 }{ 3}(1+\frac{ 1 }{ 2 }+\frac{ 1 }{ 4 }+\frac{ 1 }{ 8 }+.....)\] \[\frac{ 1 }{ 3}(\frac{ 1 }{2^{0} }+\frac{ 1 }{ 2^{1} }+\frac{ 1 }{ 2^{2} }+\frac{ 1 }{ 2^{3} }+.....)\] \[\frac{ 1 }{ 3 }(\frac{ 1\times(1-\frac{ 1 }{ 2^{n} }) }{ 1-\frac{ 1 }{ 2 } })\] \[\frac{ 1 }{ 3 }\times \frac{ 1 }{ \frac{ 1 }{ 2 } }\] \[\frac{ 2 }{ 3 }\]

  4. Yahoo!
    • 3 years ago
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    Sum to infinite G.P

  5. Yahoo!
    • 3 years ago
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    \[S \infty = \frac{ a }{ 1-r }\]

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