Question

Find the sum of the given series: $0.4+0.44+0.444+......$ to n terms?

Answer 1

Hint: We start solving the problem by taking 4 common from each term of the given series. We then multiply and divide the series with 9. We then make the necessary calculations and make use of the result that the sum of of n terms in geometric progression a, ar, $a{{r}^{2}}$, ……, $a{{r}^{n-1}}$ is $\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}$ for $\left| r \right|<1$. We then make the necessary calculations after applying the result to get the required answer.

Complete step by step answer:
According to the problem, we need to find the sum of the given series: $0.4+0.44+0.444+......$ to n terms.
Let us assume $S=0.4+0.44+0.444+......$.
$\Rightarrow S=4\left( 0.1+0.11+0.111+...... \right)$.
Let us multiply and divide R.H.S (Right Hand Side) with 9.
$\Rightarrow S=\dfrac{4}{9}\left( 0.9+0.99+0.999+...... \right)$.
$\Rightarrow S=\dfrac{4}{9}\left( \left( 1-0.1 \right)+\left( 1-0.01 \right)+\left( 1-0.001 \right)+...... \right)$.
$\Rightarrow S=\dfrac{4}{9}\left( \left( 1+1+1+......n\text{ terms} \right)-\left( 0.1+0.01+0.001+......n\text{ terms} \right) \right)$.
$\Rightarrow S=\dfrac{4}{9}\left( n-\left( \dfrac{1}{10}+\dfrac{1}{{{10}^{2}}}+\dfrac{1}{{{10}^{3}}}+......+\dfrac{1}{{{10}^{n}}} \right) \right)$ ---(1).
We can see that the sum $\dfrac{1}{10}+\dfrac{1}{{{10}^{2}}}+\dfrac{1}{{{10}^{3}}}+.....\dfrac{1}{{{10}^{n}}}$ resembles sum of ‘n’ terms in geometric progression with first term $\dfrac{1}{10}$ and common ratio $\dfrac{1}{10}$.
We know that the sum of n terms in geometric progression a, ar, $a{{r}^{2}}$, ……, $a{{r}^{n-1}}$ is $\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}$ for $\left| r \right|<1$.
So, we get $\dfrac{1}{10}+\dfrac{1}{{{10}^{2}}}+\dfrac{1}{{{10}^{3}}}+.....\dfrac{1}{{{10}^{n}}}=\dfrac{\dfrac{1}{10}\left( 1-\dfrac{1}{{{10}^{n}}} \right)}{\left( 1-\dfrac{1}{10} \right)}$.
$\Rightarrow \dfrac{1}{10}+\dfrac{1}{{{10}^{2}}}+\dfrac{1}{{{10}^{3}}}+.....\dfrac{1}{{{10}^{n}}}=\dfrac{\dfrac{1}{10}\left( 1-\dfrac{1}{{{10}^{n}}} \right)}{\dfrac{9}{10}}$.
$\Rightarrow \dfrac{1}{10}+\dfrac{1}{{{10}^{2}}}+\dfrac{1}{{{10}^{3}}}+.....\dfrac{1}{{{10}^{n}}}=\dfrac{1}{9}\left( 1-\dfrac{1}{{{10}^{n}}} \right)$ ---(2).
Let us substitute equation (2) in equation (1).
$\Rightarrow S=\dfrac{4}{9}\left( n-\dfrac{1}{9}\left( 1-\dfrac{1}{{{10}^{n}}} \right) \right)$.
$\Rightarrow S=\dfrac{4}{9}\left( \dfrac{1}{9}\left( 9n-1+\dfrac{1}{{{10}^{n}}} \right) \right)$.
$\Rightarrow S=\dfrac{4}{81}\left( 9n-1+\dfrac{1}{{{10}^{n}}} \right)$.

$\therefore $We have found the sum of the series $0.4+0.44+0.444+......$ to n terms as $\dfrac{4}{81}\left( 9n-1+\dfrac{1}{{{10}^{n}}} \right)$.

Note: We can see that the given problem contains a huge amount of calculation so, we need to perform each step carefully in order to avoid confusion and calculation mistakes. Whenever we get this type of problem, we try to convert the given series into a sum of n and a geometric progression which makes the problem easier. We should not confuse the sum of n terms of geometric progression while solving this type of problem. Similarly, we can expect problems to find the sum of the series $4+44+444+.....$ n terms.