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5 + 55 + 555 + ….

.6 + .66 + .666 + ….

Answer
Verified

Hint: First of all, multiply and divide 9 in both the series after taking common 5 and 0.6 from them respectively. Then make a G.P of the form \[10,{{10}^{2}},{{10}^{3}}....\] by writing 9 = 10 – 1, 99 = 100 – 1, and so on. Finally, use the formula for the sum of n terms of G.P that is \[\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\].

__Complete step-by-step solution -__

In this question, we have to find the sum of the n terms of the series.

(i) 5 + 55 + 555 + ….

(ii) .6 + .66 + .666 + ….

Let us consider the first series

S = 5 + 55 + 555 + ….. n terms

By taking out 5 common from the above series, we get,

S = 5 (1 + 11 + 111 + 1111 …. n terms)

By multiplying 9 on both the sides of the above equation, we get,

9S = 5 [9 + 99 + 999 …… n times]

\[\dfrac{9S}{5}=9+99+999.....\text{ n times}\]

We can also write the above equation as,

\[\dfrac{9S}{5}=\left( 10-1 \right)+\left( 100-1 \right)+\left( 1000-1 \right)....\text{ n times}\]

By simplifying the above equation and taking the term 1 separately, we get,

\[\dfrac{9S}{5}=\left( {{10}^{1}}+{{10}^{2}}+{{10}^{3}}.....\text{ n times} \right)-\left( 1+1+1......\text{ n times} \right)\]

We know that 1 added n times is equal to n and the nth term of 1, 2, 3…. n. So, we get,

\[\dfrac{9S}{5}=\left( {{10}^{1}}+{{10}^{2}}+{{10}^{3}}{{....10}^{n}} \right)-\left( n \right)\]

We know that the series of the form \[a,ar,a{{r}^{2}},a{{r}^{3}}.....a{{r}^{n}}\] is in G.P and sum of the n terms of this G.P is given by \[\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\]

In the above equation, we can see that \[10,{{10}^{2}},{{10}^{3}}....\] are in G.P with the first term and common ratio as 10. So, we get,

\[\dfrac{9S}{5}=\dfrac{10\left( 1-{{\left( 10 \right)}^{n}} \right)}{1-10}-n\]

\[\dfrac{9S}{5}=\dfrac{10\left( 1-{{\left( 10 \right)}^{n}} \right)}{-9}-n\]

\[\dfrac{9S}{5}=\dfrac{10}{9}\left( {{10}^{n}}-1 \right)-n\]

\[\dfrac{9S}{5}=\dfrac{10\left( {{10}^{n}}-1 \right)-9n}{9}\]

By multiplying \[\dfrac{5}{9}\] on both the sides of the above equation, we get,

\[S=\dfrac{5}{9}\left[ \dfrac{10\left( {{10}^{n}}-1 \right)-9n}{9} \right]\]

\[S=\dfrac{50\left( {{10}^{n}}-1 \right)}{81}-\dfrac{5}{9}n\]

Hence, the sum of the sequence 5 + 55 + 555….. up to n terms is equal to \[\dfrac{50\left( {{10}^{n}}-1 \right)}{81}-\dfrac{5n}{9}\]

Let us consider the second set.

S = 0.6 + 0.66 + 0.666 + ….. n times.

By taking out 0.6 common, we can write the sequence as,

S = 6 [0.1 + 0.11 + 0.111 ….. n terms]

By multiplying and dividing RHS by 9, we get,

\[S=\dfrac{6}{9}\left[ 0.9+0.99+0.999.....n\text{ terms} \right]\]

\[S=\dfrac{2}{3}\left[ \dfrac{9}{10}+\dfrac{99}{100}+\dfrac{999}{1000}+\dfrac{9999}{10000}.....n\text{ terms} \right]\]

We can write the above equation as,

\[S=\dfrac{2}{3}\left[ \dfrac{\left( 10-1 \right)}{10}+\dfrac{\left( 100-1 \right)}{100}+\dfrac{\left( 1000-1 \right)}{1000}+\dfrac{\left( 10000-1 \right)}{10000}.....n\text{ terms} \right]\]

\[S=\dfrac{2}{3}\left[ \left( 1-\dfrac{1}{10} \right)+\left( 1-\dfrac{1}{100} \right)+\left( 1-\dfrac{1}{1000} \right)+\left( 1-\dfrac{1}{10000} \right).....n\text{ terms} \right]\]

\[S=\dfrac{2}{3}\left[ \left( 1+1+1......n\text{ terms} \right)-\left( \dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}.+...n\text{ terms} \right) \right]\]

Here, we can see that \[\dfrac{1}{10}+\dfrac{1}{{{\left( 10 \right)}^{2}}}+\dfrac{1}{{{\left( 10 \right)}^{3}}}+.....n\text{ terms}\] is in GP with \[a=\dfrac{1}{10}\] and \[r=\dfrac{1}{10}\].

We know that the sum of n terms of the G.P is \[\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\]. So, we get,

\[S=\dfrac{2}{3}\left[ \left( n \right)-\dfrac{\dfrac{1}{10}\left( 1-{{\left( \dfrac{1}{10} \right)}^{n}} \right)}{1-\dfrac{1}{10}} \right]\]

\[S=\dfrac{2}{3}\left[ n-\dfrac{\dfrac{1}{10}\left( 1-\dfrac{1}{{{10}^{n}}} \right)}{\dfrac{9}{10}} \right]\]

\[S=\dfrac{2}{3}\left[ n-\dfrac{1}{9}\left( 1-\dfrac{1}{{{10}^{n}}} \right) \right]\]

We know that \[\dfrac{1}{{{a}^{n}}}={{a}^{-n}}\]. By using this, we get,

\[S=\dfrac{2}{3}\left( n-\dfrac{1}{9}\left( 1-{{10}^{-n}} \right) \right)\]

So, \[S=\dfrac{2}{3}n-\dfrac{2}{27}\left[ 1-{{\left( 10 \right)}^{-n}} \right]\]

Hence, we get the sum of the sequence 0.6 + 0.66 + 0.666 + ….. n terms as \[\dfrac{2}{3}n-\dfrac{2}{27}\left[ 1-{{\left( 10 \right)}^{-n}} \right]\].

Note: In this question, students must note that whenever we are in need to find the sum of series of the type a + aa + aaa + aaaa ….., we always have to make the series of 9 + 99 + 999 + ….. type by multiplying and dividing the whole series by 9 so that we can get a G.P of the type \[10,{{10}^{2}},{{10}^{3}}\]…. Also in this question, some students make this mistake of considering 5 + 55 + 555 + …. Or 0.6 + 0.66 + 0.666 + …. In G.P which is wrong because we can see that there is no common ratio in these sequences.

In this question, we have to find the sum of the n terms of the series.

(i) 5 + 55 + 555 + ….

(ii) .6 + .66 + .666 + ….

Let us consider the first series

S = 5 + 55 + 555 + ….. n terms

By taking out 5 common from the above series, we get,

S = 5 (1 + 11 + 111 + 1111 …. n terms)

By multiplying 9 on both the sides of the above equation, we get,

9S = 5 [9 + 99 + 999 …… n times]

\[\dfrac{9S}{5}=9+99+999.....\text{ n times}\]

We can also write the above equation as,

\[\dfrac{9S}{5}=\left( 10-1 \right)+\left( 100-1 \right)+\left( 1000-1 \right)....\text{ n times}\]

By simplifying the above equation and taking the term 1 separately, we get,

\[\dfrac{9S}{5}=\left( {{10}^{1}}+{{10}^{2}}+{{10}^{3}}.....\text{ n times} \right)-\left( 1+1+1......\text{ n times} \right)\]

We know that 1 added n times is equal to n and the nth term of 1, 2, 3…. n. So, we get,

\[\dfrac{9S}{5}=\left( {{10}^{1}}+{{10}^{2}}+{{10}^{3}}{{....10}^{n}} \right)-\left( n \right)\]

We know that the series of the form \[a,ar,a{{r}^{2}},a{{r}^{3}}.....a{{r}^{n}}\] is in G.P and sum of the n terms of this G.P is given by \[\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\]

In the above equation, we can see that \[10,{{10}^{2}},{{10}^{3}}....\] are in G.P with the first term and common ratio as 10. So, we get,

\[\dfrac{9S}{5}=\dfrac{10\left( 1-{{\left( 10 \right)}^{n}} \right)}{1-10}-n\]

\[\dfrac{9S}{5}=\dfrac{10\left( 1-{{\left( 10 \right)}^{n}} \right)}{-9}-n\]

\[\dfrac{9S}{5}=\dfrac{10}{9}\left( {{10}^{n}}-1 \right)-n\]

\[\dfrac{9S}{5}=\dfrac{10\left( {{10}^{n}}-1 \right)-9n}{9}\]

By multiplying \[\dfrac{5}{9}\] on both the sides of the above equation, we get,

\[S=\dfrac{5}{9}\left[ \dfrac{10\left( {{10}^{n}}-1 \right)-9n}{9} \right]\]

\[S=\dfrac{50\left( {{10}^{n}}-1 \right)}{81}-\dfrac{5}{9}n\]

Hence, the sum of the sequence 5 + 55 + 555….. up to n terms is equal to \[\dfrac{50\left( {{10}^{n}}-1 \right)}{81}-\dfrac{5n}{9}\]

Let us consider the second set.

S = 0.6 + 0.66 + 0.666 + ….. n times.

By taking out 0.6 common, we can write the sequence as,

S = 6 [0.1 + 0.11 + 0.111 ….. n terms]

By multiplying and dividing RHS by 9, we get,

\[S=\dfrac{6}{9}\left[ 0.9+0.99+0.999.....n\text{ terms} \right]\]

\[S=\dfrac{2}{3}\left[ \dfrac{9}{10}+\dfrac{99}{100}+\dfrac{999}{1000}+\dfrac{9999}{10000}.....n\text{ terms} \right]\]

We can write the above equation as,

\[S=\dfrac{2}{3}\left[ \dfrac{\left( 10-1 \right)}{10}+\dfrac{\left( 100-1 \right)}{100}+\dfrac{\left( 1000-1 \right)}{1000}+\dfrac{\left( 10000-1 \right)}{10000}.....n\text{ terms} \right]\]

\[S=\dfrac{2}{3}\left[ \left( 1-\dfrac{1}{10} \right)+\left( 1-\dfrac{1}{100} \right)+\left( 1-\dfrac{1}{1000} \right)+\left( 1-\dfrac{1}{10000} \right).....n\text{ terms} \right]\]

\[S=\dfrac{2}{3}\left[ \left( 1+1+1......n\text{ terms} \right)-\left( \dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}.+...n\text{ terms} \right) \right]\]

Here, we can see that \[\dfrac{1}{10}+\dfrac{1}{{{\left( 10 \right)}^{2}}}+\dfrac{1}{{{\left( 10 \right)}^{3}}}+.....n\text{ terms}\] is in GP with \[a=\dfrac{1}{10}\] and \[r=\dfrac{1}{10}\].

We know that the sum of n terms of the G.P is \[\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\]. So, we get,

\[S=\dfrac{2}{3}\left[ \left( n \right)-\dfrac{\dfrac{1}{10}\left( 1-{{\left( \dfrac{1}{10} \right)}^{n}} \right)}{1-\dfrac{1}{10}} \right]\]

\[S=\dfrac{2}{3}\left[ n-\dfrac{\dfrac{1}{10}\left( 1-\dfrac{1}{{{10}^{n}}} \right)}{\dfrac{9}{10}} \right]\]

\[S=\dfrac{2}{3}\left[ n-\dfrac{1}{9}\left( 1-\dfrac{1}{{{10}^{n}}} \right) \right]\]

We know that \[\dfrac{1}{{{a}^{n}}}={{a}^{-n}}\]. By using this, we get,

\[S=\dfrac{2}{3}\left( n-\dfrac{1}{9}\left( 1-{{10}^{-n}} \right) \right)\]

So, \[S=\dfrac{2}{3}n-\dfrac{2}{27}\left[ 1-{{\left( 10 \right)}^{-n}} \right]\]

Hence, we get the sum of the sequence 0.6 + 0.66 + 0.666 + ….. n terms as \[\dfrac{2}{3}n-\dfrac{2}{27}\left[ 1-{{\left( 10 \right)}^{-n}} \right]\].

Note: In this question, students must note that whenever we are in need to find the sum of series of the type a + aa + aaa + aaaa ….., we always have to make the series of 9 + 99 + 999 + ….. type by multiplying and dividing the whole series by 9 so that we can get a G.P of the type \[10,{{10}^{2}},{{10}^{3}}\]…. Also in this question, some students make this mistake of considering 5 + 55 + 555 + …. Or 0.6 + 0.66 + 0.666 + …. In G.P which is wrong because we can see that there is no common ratio in these sequences.

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