Question

Find the sum of the following series: $5+55+555+......{{n}^{th}}term$.

Answer 1

We first try to convert the given series into a series of G.P. We take common terms to make them change the digit from 5 to 9. We then convert them to their nearest power of 10 to form the G.P. We find the sum of the G.P. which helps to find the solution to the problem.

Complete step-by-step solution:
The given series of $5+55+555+......{{n}^{th}}term$ is a sum of complex G.P. series where we have to first convert the series into a G.P first.
The series is a sum of terms of an increasing number of digits in each term and the only digit used in this case is 5. We first convert that digit from 5 to 9.
We take $\dfrac{5}{9}$ common out of the sum series.
So, $5+55+555+......{{n}^{th}}term=\dfrac{5}{9}\left( 9+99+999+......{{n}^{th}}term \right)$.
Now we convert each of the terms to its immediate next term minus 1.
The immediate next term will be the lowest term of that number of digits
Like 9 changes to $\left( 10-1 \right)$ and 999 changes to $\left( 1000-1 \right)$.
The sum becomes
$\begin{aligned} & \dfrac{5}{9}\left( 9+99+999+......{{n}^{th}}term \right) \\\ & \Rightarrow \dfrac{5}{9}\left[ \left( 10-1 \right)+\left( 100-1 \right)+\left( 1000-1 \right)+......{{n}^{th}}term \right] \\\ \end{aligned}$
We take the power of 10 together and the series of 1 together.
Here we also need to define how many 1 is there. We are trying to go to n terms of the series $5+55+555+......{{n}^{th}}term$ and that means there will n numbers of ones. The same goes for the power of tens.
The equation becomes
$\begin{aligned} & \dfrac{5}{9}\left[ \left( 10-1 \right)+\left( 100-1 \right)+\left( 1000-1 \right)+......{{n}^{th}}term \right] \\\ & \Rightarrow \dfrac{5}{9}\left[ \left( 10+100+1000+......{{n}^{th}}term \right)-n \right] \\\ & \Rightarrow \dfrac{5}{9}\left[ \left( {{10}^{1}}+{{10}^{2}}+{{10}^{3}}+......+{{10}^{n}} \right)-n \right] \\\ \end{aligned}$
Now we have a G.P. which gives
$\dfrac{5}{9}\left[ \left( {{10}^{1}}+{{10}^{2}}+{{10}^{3}}+......+{{10}^{n}} \right)-n \right]=\dfrac{5}{9}\left[ 10\dfrac{{{10}^{n}}-1}{10-1}-n \right]=\dfrac{50}{81}\left( {{10}^{n}}-1 \right)-\dfrac{5n}{9}$.
Therefore, the sum of the following series: $5+55+555+......{{n}^{th}}term$ is $\dfrac{50}{81}\left( {{10}^{n}}-1 \right)-\dfrac{5n}{9}$.

Note: To find the sum of the G.P. we used the formula of $a\dfrac{{{r}^{n}}-1}{r-1}$ for the series $a,ar,a{{r}^{2}},........,a{{r}^{n-1}}$. The specific way of taking common is fixed for any digit other than 5. The main goal is to convert them to the powers of 10.