Question

Find the sum of the following series:
$0.6+0.66+0.666+.......\text{n terms}$

Answer 1

Hint: First of all take 6 as common from the given series then multiply and divide the remaining terms by 9 then you will get the terms in the bracket as 0.9, 0.99 then we can write 0.9 as $\left( 1-\dfrac{1}{10} \right)$. Similarly, you can write the other terms of the bracket too. After that, you will find that some of the terms are forming G.P. and so find the sum of the G.P. using the formula for sum of n terms of G.P. as ${{S}_{n}}=a\left( \dfrac{{{r}^{n}}-1}{r-1} \right)$ where “a” is the first term and “r” is the common ratio.

Complete step-by-step answer:
We have to find the sum of the following sequence:
$0.6+0.66+0.666+.......\text{n terms}$
Taking 6 as common from the above series we get,
$6\left( 0.1+0.11+0.111+......\text{n terms} \right)$
Now, multiplying and dividing the above expression by 9 we get,
$\dfrac{6}{9}\left( 0.9+0.99+0.999+......\text{n terms} \right)$
Now, removing the decimal in the above series like writing 0.9 as $\dfrac{9}{10}$ we get,
$\dfrac{6}{9}\left( \dfrac{9}{10}+\dfrac{99}{100}+\dfrac{999}{1000}+......\text{n terms} \right)$
In the above series, we can write:
$\begin{align}
  & \dfrac{9}{10}=1-\dfrac{1}{10}, \\
 & \dfrac{99}{100}=1-\dfrac{1}{100} \\
\end{align}$
Likewise you can write the other terms too.
$\dfrac{6}{9}\left( 1-\dfrac{1}{10}+1-\dfrac{1}{100}+1-\dfrac{1}{1000}+......\text{n terms} \right)$
Now, in the above series, 1 has been written n times so the summation of all 1 in the above series is n.
\[\begin{align}
  & \dfrac{6}{9}\left( \left( 1+1+1.....\text{n terms} \right)-\left( \dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}+......\text{n terms} \right) \right) \\
 & =\dfrac{6}{9}\left( n-\left( \dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}+......\text{n terms} \right) \right).........Eq.(1) \\
\end{align}\]
From the above series,
\[\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}+......\text{n terms}\]
The above series is in G.P. in which the common ratio is $\dfrac{1}{10}$ and the first term is $\dfrac{1}{10}$.
Common ratio we have calculated by dividing the second term by first term. Let us take any term say $\dfrac{1}{100}$ and dividing this term by $\dfrac{1}{10}$ we get,
$\dfrac{\dfrac{1}{100}}{\dfrac{1}{10}}=\dfrac{1}{10}$
We know that, sum of n terms in G.P. is equal to:
${{S}_{n}}=a\left( \dfrac{{{r}^{n}}-1}{r-1} \right)$
In the above formula, “a” is the first term and “r” is the common ratio so substituting the value of “a” as $\dfrac{1}{10}$ and “r” as $\dfrac{1}{10}$ we get,
$\begin{align}
  & {{S}_{n}}=\dfrac{1}{10}\left( \dfrac{{{\left( \dfrac{1}{10} \right)}^{n}}-1}{\dfrac{1}{10}-1} \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{1}{10}\left( \dfrac{\dfrac{1-{{\left( 10 \right)}^{n}}}{{{\left( 10 \right)}^{n}}}}{\dfrac{-9}{10}} \right) \\
 & \Rightarrow {{S}_{n}}=-\dfrac{1}{9}\left( \dfrac{1-{{\left( 10 \right)}^{n}}}{{{\left( 10 \right)}^{n}}} \right) \\
\end{align}$
Rearranging the above equation we get,
$\begin{align}
  & {{S}_{n}}=-\dfrac{1}{9}\left( \dfrac{1}{{{\left( 10 \right)}^{n}}}-1 \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{1}{9}\left( 1-\dfrac{1}{{{\left( 10 \right)}^{n}}} \right) \\
\end{align}$
Substituting the above summation in eq. (1) we get,
\[\dfrac{6}{9}\left( n-\dfrac{1}{9}\left( 1-\dfrac{1}{{{\left( 10 \right)}^{n}}} \right) \right)\]
Hence, we got the summation of the given series as:
\[\dfrac{6}{9}\left( n-\dfrac{1}{9}\left( 1-\dfrac{1}{{{\left( 10 \right)}^{n}}} \right) \right)\]

Note: You might have confused the number of terms when we converted the following:
$\begin{align}
  & \dfrac{9}{10}=1-\dfrac{1}{10}, \\
 & \dfrac{99}{100}=1-\dfrac{1}{100} \\
\end{align}$
Which we written in the above solution as:
$\dfrac{6}{9}\left( \dfrac{9}{10}+\dfrac{99}{100}+\dfrac{999}{1000}+......\text{n terms} \right)$
$\dfrac{6}{9}\left( 1-\dfrac{1}{10}+1-\dfrac{1}{100}+1-\dfrac{1}{1000}+......\text{n terms} \right)$
As you can see that we have converted each term written in the bracket and as there are n terms in the bracket so the converted terms are also n and there are n number of ones and the remaining terms are also n.