The sum of first n terms of the series 4/3,10/9,28/27,82/81,244/243, is

Correct Answer is : n+12(1−3−n)n+1/2(1-3^{-n})n+21(1−3−n)

Correct Answer is : n+12(1−3−n)n+1/2(1-3^{-n})n+21(1−3−n)

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KEAM 2012 Math Paper

Question : 37 of 120

Marks: +1, -0

n

\frac{4}{3},\frac{10}{9},\frac{28}{27},\frac{82}{81},\frac{244}{243},\dots

$n+\frac{1}{2}(1+3^{-n})$
$n-\frac{1}{2}(1+3^{-n})$
$n+\frac{1}{2}(2+3^{-n})$
$n+\frac{1}{2}(2-3^{-n})$
$n+\frac{1}{2}(1-3^{-n})$

Solution:

Given,

n

series is

\;\frac{4}{3},\;\frac{10}{9},\;\frac{28}{27},\;\frac{82}{81},\;\frac{244}{243},\dots

Here,

S_{1}=\;\frac{4}{3},\;\;S_{2}=\;\frac{4}{3}+\;\frac{10}{9}=\;\frac{12+10}{9}=\;\frac{22}{9}

Now, taking option (e)

S_{n}=n+\;\frac{1}{2}(1-3^{-n})

Put

\;\;n=1

S_{1}=1+\;\frac{1}{2}\left(1-\;\frac{1}{3}\right)=1+\;\frac{1}{2}\left(\;\frac{2}{3}\right)

=\;\frac{4}{3},

which is true.

Put

n=2

S_{2}=2+\;\frac{1}{2}\left(1-\;\frac{1}{3^{2}}\right)

=2+\;\frac{1}{2}\left(\;\frac{8}{9}\right)=2+\;\frac{4}{9}

=\;\frac{22}{9},

which is true.

Hence, option (e) is true.

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