Question

The sum of $n$ terms of the series $1+\dfrac{4}{5}+\dfrac{7}{{{5}^{2}}}+\dfrac{10}{{{5}^{3}}}+...$ is A. $\dfrac{5}{4}+\dfrac{15}{16}\left( 1-\dfrac{1}{{{5}^{n-1}}} \right)-\dfrac{\left( 3n-2 \right)}{4\cdot {{5}^{n-1}}}$$$$$ B. $\dfrac{5}{4}+\dfrac{15}{16}\left( 1-\dfrac{1}{{{5}^{n-1}}} \right)-\dfrac{3n}{4\cdot {{5}^{n-1}}}$$$$$ C. $\left( 1-\dfrac{1}{{{5}^{n-1}}} \right)-\dfrac{\left( 3n+2 \right)}{4\cdot {{5}^{n-1}}}$$$$$ D. None of these

Answer 1

We find the ${{n}^{\text{th}}}$ term of the series by observing the pattern as $\dfrac{3n-2}{{{5}^{n-1}}}$. We denote the sum up to $n$ terms as $S=1+\dfrac{4}{5}+\dfrac{7}{{{5}^{2}}}+\cdots$and then divide both side of the equation by 5 to get $\dfrac{S}{5}$. We subtract $S-\dfrac{S}{5}$ , take 3 common and find GP series with first term $a=\dfrac{1}{5}$ and common ratio $r=\dfrac{1}{5}$. We use the sum of $n$ terms in GP series formula $\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$ and simplify to get an expression in $n$.$$$$

Complete step-by-step solution:
We know that GP series is the sum of terms of GP sequence where the consecutive terms are in constant ratio. The sum of first $n$ terms with first term $a$ and common ratio $r$ is given by
$\text{sum}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$
We are given in the question the series with infinite terms as,
$1+\dfrac{4}{5}+\dfrac{7}{{{5}^{2}}}+\dfrac{10}{{{5}^{3}}}+...\text{ }$
We observe the pattern of the terms and see that ${{n}^{\text{th}}}$ terms of the above sequence is $\dfrac{3n-2}{{{5}^{n-1}}}$and hence the ${{\left( n-1 \right)}^{\text{th}}}$ term of the series is $\dfrac{3\left( n-1 \right)-2}{{{5}^{\left( n-1 \right)-1}}}=\dfrac{3n-5}{{{5}^{n-2}}}$ So we can write the series as up to ${{n}^{\text{th}}}$ term as and denote the sum of $n$ terms as $S$ to have,
$S=1+\dfrac{4}{5}+\dfrac{7}{{{5}^{2}}}+\dfrac{10}{{{5}^{3}}}+\cdots \text{+}\dfrac{3n-5}{{{5}^{n-2}}}+\dfrac{3n-2}{{{5}^{n-1}}}\text{ }...\left( 1 \right)$
Let us divide the above series both side of the above equation by 5 and have,
$\dfrac{S}{5}=\dfrac{1}{5}+\dfrac{4}{{{5}^{2}}}+\dfrac{7}{{{5}^{3}}}+\dfrac{10}{{{5}^{4}}}+\cdots \text{+}\dfrac{3n-5}{{{5}^{n-1}}}+\dfrac{3n-2}{{{5}^{n}}}\text{ }....\left( 2 \right)$
Let us subtract equation (2) from equation (1) and have,

& S-\dfrac{S}{5}=\left( 1+\dfrac{4}{5}+\dfrac{7}{{{5}^{2}}}+\dfrac{10}{{{5}^{3}}}+\cdots \text{+}\dfrac{3n-5}{{{5}^{n-2}}}+\dfrac{3n-2}{{{5}^{n-1}}}\text{ } \right) \\\ & -\left( \dfrac{1}{5}+\dfrac{4}{{{5}^{2}}}+\dfrac{7}{{{5}^{3}}}+\dfrac{10}{{{5}^{4}}}\cdots \text{+}\dfrac{3n-5}{{{5}^{n-1}}}+\dfrac{3n-2}{{{5}^{n}}} \right) \\\ & \Rightarrow \dfrac{5S-S}{5}=\left( 1+\dfrac{4}{5}+\dfrac{7}{{{5}^{2}}}+\dfrac{10}{{{5}^{3}}}+\dfrac{13}{{{5}^{4}}}\cdots \text{+}\dfrac{3n-5}{{{5}^{n-2}}}+\dfrac{3n-2}{{{5}^{n-1}}} \right) \\\ & -\dfrac{1}{5}-\dfrac{4}{{{5}^{2}}}-\dfrac{7}{{{5}^{3}}}-\dfrac{10}{{{5}^{4}}}\cdots -\dfrac{3n-5}{{{5}^{n-1}}}-\dfrac{3n-2}{{{5}^{n}}} \\\ \end{aligned}$$ Let us write the terms with same denominator close to each other.We have, $$\begin{aligned} & \dfrac{4S}{5}=1+\dfrac{4}{5}-\dfrac{1}{5}+\dfrac{7}{{{5}^{2}}}-\dfrac{4}{{{5}^{2}}}+\dfrac{10}{{{5}^{3}}}-\dfrac{7}{{{5}^{3}}}+\ldots +\dfrac{3n-2}{{{5}^{n-1}}}-\dfrac{3n-5}{{{5}^{n-1}}}-\dfrac{3n-2}{{{5}^{n-1}}} \\\ & \Rightarrow \dfrac{4S}{5}=1+\dfrac{4-1}{5}+\dfrac{7-4}{{{5}^{2}}}+\dfrac{10-7}{{{5}^{3}}}+\ldots +\dfrac{3n-2-\left( 3n-5 \right)}{{{5}^{n-1}}}-\dfrac{3n-2}{{{5}^{n-1}}} \\\ & \Rightarrow \dfrac{4S}{5}=1+\dfrac{3}{5}+\dfrac{3}{{{5}^{2}}}+\dfrac{3}{{{5}^{3}}}+\ldots +\dfrac{3}{{{5}^{n-1}}}-\dfrac{3n-1}{{{5}^{n-1}}} \\\ \end{aligned}$$ We see in above step that first term 1 and last term $-\dfrac{3n-1}{{{5}^{n-1}}}$ are unique terms and we take 3 common from rest of the terms to have, $$\Rightarrow \dfrac{4S}{5}=1+3\left( \dfrac{1}{5}+\dfrac{1}{{{5}^{2}}}+\dfrac{1}{{{5}^{3}}}+\ldots +\dfrac{1}{{{5}^{n-1}}} \right)-\dfrac{3n-1}{{{5}^{n-1}}}$$ We see in above step that the series inside the bracket is a GP series with $n-1$ terms where first term is $a=\dfrac{1}{5}$ and common ratio $r=\dfrac{1}{{{5}^{2}}}\div \dfrac{1}{5}=\dfrac{1}{5}$. We use the sum of finite terms and have, $$\begin{aligned} & \Rightarrow \dfrac{4S}{5}=1+3\times \dfrac{1}{5}\left( \dfrac{1-{{\left( \dfrac{1}{5} \right)}^{n-1}}}{1-\dfrac{1}{5}} \right)-\dfrac{3n-1}{{{5}^{n-1}}} \\\ & \Rightarrow \dfrac{4S}{5}=1+\dfrac{3}{5}\left( \dfrac{1-\dfrac{1}{{{5}^{n-1}}}}{\dfrac{4}{5}} \right)-\dfrac{3n-1}{{{5}^{n-1}}} \\\ & \Rightarrow \dfrac{4S}{5}=1+\dfrac{15}{20}\left( 1-\dfrac{1}{{{5}^{n-1}}} \right)-\dfrac{3n-1}{{{5}^{n-1}}} \\\ \end{aligned}$$ We multiply $\dfrac{5}{4}$ both side of the equation in the above step to have, $$\begin{aligned} & \Rightarrow \dfrac{4S}{5}\times \dfrac{5}{4}=1\times \dfrac{5}{4}+\dfrac{15}{20}\times \dfrac{5}{4}\left( 1-\dfrac{1}{{{5}^{n-1}}} \right)-\dfrac{3n-2}{{{5}^{n-1}}}\times \dfrac{5}{4} \\\ & \Rightarrow S=\dfrac{5}{4}+\dfrac{15}{16}\times \left( 1-\dfrac{1}{{{5}^{n-1}}} \right)-\dfrac{3n-2}{4\cdot {{5}^{n-2}}} \\\ \end{aligned}$$ **The above expression is the required expression for sum of $n$ terms of the series and the correct option A.** **Note:** We see that in the numerator terms of the given series $1+\dfrac{4}{5}+\dfrac{7}{{{5}^{2}}}+\dfrac{10}{{{5}^{3}}}+...$ are in AP sequence and the denominators are in GP series and that is why we used this strategy. If the question of would have asked for sum with infinite terms we would use the sum of terms in a GP sequence $S=\dfrac{a}{1-r}$ with condition $\left| r \right | < 1 $ which in this problem is satisfied as $r=\dfrac{1}{5}$.