Question

Find the sum of n terms of the sequence: $ 5 + 55 + 555 + ........ $

Answer 1

Hint: As we can see every term is a multiple of 5. So take out 5 common. And then multiply and divide each term with 9. Now after taking 9 common in the denominator of each term out we will have a series in the form $ \dfrac{5}{9}\left( {9 + 99 + 999 + .....} \right) $ . 9 can be written as $ \left( {10 - 1} \right) $ , 99 can be written as $ \left( {100 - 1} \right) $ and so on. Finally we will get one geometric series and one series with only 1s which are added n times. Find the sum of n terms of a geometric series using the below formula to get the required sum.
Formula used:
Sum of n terms of a geometric sequence is $ \dfrac{{a\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}} $ , where a is the first term and r is the common ratio.

Complete step-by-step answer:
We are given to find the sum of n terms of the sequence $ 5 + 55 + 555 + ........ $
As we can see 5 is a multiple of 5, 55 is a multiple of 5, 555 is a multiple of 5 and so on.
So on taking out 5 common from the sequence, we get
$\Rightarrow 5\left( {1 + 11 + 111 + ....} \right) $
Now on multiplying and dividing each term inside the bracket by 9, we get
$\Rightarrow 5\left( {\dfrac{9}{9} + \dfrac{{99}}{9} + \dfrac{{999}}{9} + ....} \right) $
Now on taking out 9 common from the denominator of each term, we get
$\Rightarrow \dfrac{5}{9}\left( {9 + 99 + 999 + .....} \right) $
We know that 9 can be written as $ \left( {10 - 1} \right) $ , 99 can be written as $ \left( {100 - 1} \right) $ , 999 can be written as $ \left( {1000 - 1} \right) $ and so on.
So on replacing the terms of $ \dfrac{5}{9}\left( {9 + 99 + 999 + .....} \right) $ , we get
$\Rightarrow \dfrac{5}{9}\left[ {\left( {10 - 1} \right) + \left( {100 - 1} \right) + \left( {1000 - 1} \right) + .....} \right] $
Putting all the multiples of 10 one side and 1s another side, we get
$ \dfrac{5}{9}\left[ {\left( {10 + 100 + 1000 + .....ntimes} \right) - \left( {1 + 1 + 1 + ...ntimes} \right)} \right] $
As we can see the sequence $ \left( {10 + 100 + 1000 + .....ntimes} \right) $ is a geometric sequence with 10 as first term and 10 as common ratio.
Sum of n terms of a geometric sequence with 10 as first term and 10 as common ratio is
$ \dfrac{{10\left( {{{10}^n} - 1} \right)}}{{\left( {10 - 1} \right)}} $
Replace the series $ \left( {10 + 100 + 1000 + .....ntimes} \right) $ with $ \dfrac{{10\left( {{{10}^n} - 1} \right)}}{{\left( {10 - 1} \right)}} $
Therefore, we get
$ \dfrac{5}{9}\left[ {\left( {\dfrac{{10\left( {{{10}^n} - 1} \right)}}{{\left( {10 - 1} \right)}}} \right) - \left( {1 + 1 + 1 + ....ntimes} \right)} \right] $
Adding 1 ‘n’ times gives ‘n’.
 $ \Rightarrow \dfrac{5}{9}\left[ {\left( {\dfrac{{10\left( {{{10}^n} - 1} \right)}}{9}} \right) - n} \right] $
Now separating the above subtraction gives $ \dfrac{{50}}{{81}}\left( {{{10}^n} - 1} \right) - \dfrac{{5n}}{9} $
Therefore, the sum of n terms of the sequence $ 5 + 55 + 555 + ........ $ is $ \dfrac{{50}}{{81}}\left( {{{10}^n} - 1} \right) - \dfrac{{5n}}{9} $
So, the correct answer is “ $ \dfrac{{50}}{{81}}\left( {{{10}^n} - 1} \right) - \dfrac{{5n}}{9} $ ”.

Note: We have considered the sequence $ \left( {10 + 100 + 1000 + .....ntimes} \right) $ is in G.P because in a G.P every term starting from the second term is obtained by multiplying its previous term with a fixed ratio. This fixed ratio is called the common ratio. Do not confuse a G.P with an A.P