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\begin{align} S_n &=\frac 1 {1-1/5}+\frac{ 3(\frac{1}{5})[1-(\frac{1}{5}){^n}^{-1}} {(1- 1/5)^2}-\frac {[1+(n-1)(3)](\frac 15)^n} {1-1/5} \\ &= \frac 54+\frac {15} {16}[1-(1/5){^n}^{-1}]-\frac54[3n-2](1/5){^n}\\ &= \frac 54+\frac {15} {16}-\left[\frac {15} {16}+\frac{3n-2}{4}\right]\left(\frac 15\right){^n}^{-1} \end{align}

How this came in the 3rd line from the 2nd line. Can anyone explain this without skipping any step. $$\left[\frac {15} {16}+\frac{3n-2}{4}\right]\left(\frac 15\right){^n}^{-1}$$

2 Answers 2

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\begin{align} S_n &=\frac 1 {1-1/5}+\frac{ 3(\frac{1}{5})[1-(\frac{1}{5}){^n}^{-1}} {(1- 1/5)^2}-\frac {[1+(n-1)(3)](\frac 15)^n} {1-1/5} \\ &= \frac 54+\frac {15} {16}[1-(1/5){^n}^{-1}]-\frac54[3n-2](1/5){^n}\\ &=\frac 54+\frac {15} {16}-\frac {15} {16}(1/5){^n}^{-1}-\frac{15n}{4}(1/5){^n}+\frac{10}{4}(1/5){^n}\\ &=\frac 54+\frac {15} {16}-\frac {15} {16}(1/5){^n}^{-1}-\frac{15n}{4}(1/5){^n}^{-1}\frac 15+\frac{10}{4}(1/5){^n}^{-1}\frac 15\\ &=\frac 54+\frac {15} {16}-(1/5){^n}^{-1}\left[\frac{15}{16}+\frac{15n}{20}-\frac{10}{20}\right]\\ &=\frac 54+\frac {15} {16}-(1/5){^n}^{-1}\left[\frac{15}{16}+\frac{3n}{4}-\frac{1}{2}\right]\\ &= \frac 54+\frac {15} {16}-\left[\frac {15} {16}+\frac{3n-2}{4}\right]\left(\frac 15\right){^n}^{-1} \end{align}

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    I understand every thing but please explain me you multiplied $\frac {15n}{4}$ with $(1/5){^n}$ in the 3rd line, but where does the power n goes. – 2017-01-16
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    I take $\frac{1}{5^{n-1}}$ as common term ,Then leave with only $\frac 15$ and then multiplied$\frac{15n}{4}$ with $\frac 15$ to get $\frac {15n}{20}$ – 2017-01-16
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    sorry but will you please add one or two line for easy understanding. – 2017-01-16
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    check the edited answer. – 2017-01-16
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    yes i got it now. – 2017-01-16
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    By expanding completely, you make it more complicated than necessary. – 2017-01-16
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    @YvesDaoust But the OP just wanted to understand the expanded answer. – 2017-01-16
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    @MatheMagic: yep, you made the explanation more complicated than necessary. – 2017-01-16
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    @ YvesDaoust:Intially my answer was quite simple but as the OP requested ,i just edited it. – 2017-01-16
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    @MatheMagic Splitting the factor $(3n-2)$ and simplifying the fractions all the time is counterproductive as you have to undo later. – 2017-01-16
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    @Yves Daoust and MatheMagic I think MatheMagic explained the answer in very easy way which was helpful to me and i think it will be helpful to others also because expansion doesn't matter. If we can't understand anything after posting in math stack the website will be of no use, So please try to understand, everyone is not brilliant therefore they may need elaborated answer. Lines can be skipped after understanding. – 2017-01-16
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    @bappy: sorry, you do not convince me. This explanation is more complicated. – 2017-01-16
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$$\frac 54+\frac {15} {16}\left[1-\left(\frac15\right)^{n-1}\right]-\frac54(3n-2)\left(\frac15\right)^n=$$ Distribute $\color{green}{15/16}$ and extract a factor $\color{blue}{1/5}$, $$\frac 54+\color{green}{\frac {15} {16}}-\color{green}{\frac {15} {16}}\left(\frac15\right)^{n-1}-\frac54\color{blue}{\frac15}(3n-2)\left(\frac15\right)^{\color{blue}{n-1}}=$$ Factor out $\color{green}{(1/5)^{n-1}}$, $$\frac 54+\frac {15} {16}-\left[\frac {15} {16}+\frac14(3n-2)\right]\color{green}{\left(\frac15\right)^{n-1}}.$$