Question

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

Answer 1

\[[\sqrt{(3n)^3} -2]/(9n-3)\]

Answer 2

its wrong

Answer 3

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

Answer 4

Sum to infinite G.P

Answer 5

\[S \infty = \frac{ a }{ 1-r }\]