Answer

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**Hint:**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.

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