Question

Find the sum of the following series up to n terms: 0.6+0.66+0.666+……?

Answer 1

This type of problem is based on the concept of geometric series. First, we have to consider the sum of n terms of the given series to be ${{S}_{n}}$. Now, take 6 common from the series. Multiply and divide the whole series by 9. Keep $\dfrac{6}{9}$ outside the bracket and multiply 9 in the numerator with each term in the series. Substitute $0.9=1-\dfrac{1}{10}$, $0.99=1-\dfrac{1}{100}$, $0.999=1-\dfrac{1}{1000}$ and so on for n terms. Since there is 1 n times in the series, group all the fractions separately. We see a geometric series in the expression. Assume $\dfrac{1}{10}$ to be ‘a’. Then to find the common ratio ‘r’ of the given geometric series, we have to divide the first term and the second term of the series. We get $r=\dfrac{1}{10}$. And substitute these values in the formula to find the sum of n terms of series of a geometric progression, that is, $S=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$

Complete step by step solution:
According to the question, we are asked to find the sum of the series 0.6+0.66+0.666+….up to n terms.
We have been given the series is 0.6, 0.66, 0.666… -----(1)
Let us assume ${{S}_{n}}$ to be the sum of n terms of the given series.
$\Rightarrow {{S}_{n}}=0.6+0.66+0.666+.....$n terms.
We find that 6 are common in the RHS.
Let us take 6 outside the bracket.
$\Rightarrow {{S}_{n}}=6\left( 0.1+0.11+0.111+..... \right)$
Now, we need to multiply 9 in the numerator and denominator of the RHS.
We get
${{S}_{n}}=6\times \dfrac{9}{9}\left( 0.1+0.11+0.111+..... \right)$n terms.
Use distributive property, that is $a\left( b+c+d+.... \right)=ab+ac+ad+...$, in the numerator of the RHS.
We get
${{S}_{n}}=\dfrac{6}{9}\left( 0.9+0.99+0.999+..... \right)$n terms. ---------(2)
We know 0.9=1-0.1, but we can write 0.1 as $\dfrac{1}{10}$.
Thus, $0.9=1-\dfrac{1}{10}$.
Now, we can write 0.99 as 1-0.01 which is $1-\dfrac{1}{100}$.
And we get $0.99=1-\dfrac{1}{100}$.
Similarly, we get $0.999=1-\dfrac{1}{1000}$ and we can find the same for n terms.
Let us substitute these values in equation (2).
$\Rightarrow {{S}_{n}}=\dfrac{6}{9}\left( 1-\dfrac{1}{10}+1-\dfrac{1}{100}+1-\dfrac{1}{1000}+..... \right)$ up to n terms.
We find that 1 is added n times in the RHS. Group all of them together and we get n in the RHS.
Therefore, we get
${{S}_{n}}=\dfrac{6}{9}\left( n-\dfrac{1}{10}-\dfrac{1}{100}-\dfrac{1}{1000}-..... \right)$ up to n terms.
Now, let us take the negative sign outside the bracket.
$\Rightarrow {{S}_{n}}=\dfrac{6}{9}\left( n-\left( \dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}+..... \right) \right)$ up to n terms.
Consider $\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}+....$n terms.
We find that the series (3) is a geometric series.
Therefore, the first term is $a=\dfrac{1}{10}$.
We need to find the common ratio.
Divide the second term of the series by the first term to find the common ratio r.
$\Rightarrow r=\dfrac{\dfrac{1}{100}}{\dfrac{1}{10}}$
Using the rule $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}$ in the above expression, we get
$r=\dfrac{1}{100}\times \dfrac{10}{1}$
$\Rightarrow r=\dfrac{1}{10\times 10}\times \dfrac{10}{1}$
Here, we find that 10 are common in both the numerator and denominator. Cancelling 10, we get
$r=\dfrac{1}{10}$
We know that the formula to find the sum of n terms in a geometric series is $S=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$.
Substituting the values of ‘a’ and ‘r’, we get
$S=\dfrac{\dfrac{1}{10}\left( 1-{{\left( \dfrac{1}{10} \right)}^{n}} \right)}{1-\dfrac{1}{10}}$
Take LCM in the denominator. We get
$S=\dfrac{\dfrac{1}{10}\left( 1-{{\left( \dfrac{1}{10} \right)}^{n}} \right)}{\dfrac{10-1}{10}}$
$\Rightarrow S=\dfrac{\dfrac{1}{10}\left( 1-{{\left( \dfrac{1}{10} \right)}^{n}} \right)}{\dfrac{9}{10}}$
We find that $\dfrac{1}{10}$ is common in both the numerator and the denominator. Cancelling $\dfrac{1}{10}$, we get
$S=\dfrac{1}{9}\left( 1-{{\left( \dfrac{1}{10} \right)}^{n}} \right)$
Let us use the property ${{\left( \dfrac{a}{b} \right)}^{n}}=\dfrac{{{a}^{n}}}{{{b}^{n}}}\times \dfrac{d}{c}$ to simplify S further.
$\Rightarrow S=\dfrac{1}{9}\left( 1-\dfrac{{{1}^{n}}}{{{10}^{n}}} \right)$
Since 1 power any term is always 1, we get
$S=\dfrac{1}{9}\left( 1-\dfrac{1}{{{10}^{n}}} \right)$
Now substitute the value of ‘S’ in equation (3).
${{S}_{n}}=\dfrac{6}{9}\left[ n-\dfrac{1}{9}\left( 1-\dfrac{1}{{{10}^{n}}} \right) \right]$

Therefore, the sum of n terms of the series 0.6+0.66+0.666+….. is $\dfrac{6}{9}\left[ n-\dfrac{1}{9}\left( 1-\dfrac{1}{{{10}^{n}}} \right) \right]$.

Note: Whenever you get this type of problems, we should always try to make the necessary calculations in the given series to convert the series into a geometric or arithmetic series. We should avoid calculation mistakes based on sign conventions. If we get r>0, we have to use the formula $S=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$. Also we should not use the formula $S=\dfrac{a}{r-1}$ to find the sum of n terms. This formula is used to find the sum of infinite series.