Question

Find the sum of \[5{\text{ + 55 + 555 + 5555 + }}......{\text{ n terms}}\]

Answer 1

Hint: Given series is the sum of sequences which lasts up to n terms and therefore infinite series. There are two types of sequences and series. 1) Arithmetic progression and 2) Geometric Progression. In arithmetic progression, the difference between the numbers is constant in the series whereas the geometric progression is the sequence in which the succeeding element is obtained by multiplying the preceding number by the constant and the same continues for the series. The ratio between the two remains the same.

Complete step by step solution:
General form of geometric progression is –
$ {\text{Sum = a + ar + a}}{{\text{r}}^2} + a{r^3} + .....{\text{ + a}}{{\text{r}}^{n - 1}} $
Here For geometric Progression use –
$ {S_n} = a(\dfrac{{{r^n} - 1}}{{r - 1}}),\;{\text{r > 1}} $ where, “a” is the first term and “r” is the ratio.
Complete Step by Step solution –
\[5{\text{ + 55 + 555 + 5555 + }}......{\text{ n terms}}\]
Take 5 common from the series –
$ \Rightarrow 5{\text{ (1 + 11 + 111 + }}......{\text{ n terms)}} $
Multiply and divide by $9$
$ \Rightarrow \dfrac{5}{9}(9{\text{ + 99 + 999 + }}......{\text{)}} $
Write, $10 - 1 = 9$ in the above equation –
$
   \Rightarrow \dfrac{5}{9}[(10 - 1) + (100 - 1) + (1000 - 1) + ....{\text{ n terms ]}} \\
   \Rightarrow \dfrac{5}{9}[(10 + 100 + 1000 + .....n{\text{ terms) - (1 + 1 + 1 + }}...{\text{ n terms)]}} \\
 $
 [Taking all one’s together so being number one n terms we can write it as simply “n”]
$ \Rightarrow \dfrac{5}{9}[(10 + 100 + 1000 + .....) - n]$ ................. Eq. (1)
Now, checking ratios of two elements - $ \dfrac{{100}}{{10}} = \dfrac{{1000}}{{100}} = 10 $
So, the given sequence is of the geometric progression with ratio, $ r = 10 $ and $ a = 10 $
Since,
      $
  {\text{r > 1}} \\
  \therefore {S_n} = a(\dfrac{{{r^n} - 1}}{{r - 1}}),\; \\
 $
Put values, in Eq. (1)
$
   \Rightarrow \dfrac{5}{9}[\dfrac{{10({{10}^n} - 1)}}{{(10 - 1)}} - n] \\
   \Rightarrow \dfrac{5}{9}[\dfrac{{10({{10}^n} - 1)}}{9} - n] \\
   \Rightarrow \dfrac{5}{9}[\dfrac{{10({{10}^n} - 1) - 9n}}{9}] \\
   \Rightarrow \dfrac{{50}}{{81}}[({10^n} - 1) - 9n] \\
   \Rightarrow \dfrac{{50}}{{81}}[{10^n} - 9n - 1] \\
 $
The required solution is -
$ 5{\text{ + 55 + 555 + }}....{\text{ n terms = }}\dfrac{{50}}{{81}}[{10^n} - 9n - 1] $

Additional Information: The arithmetic series is the sum of all the terms of the arithmetic progression (AP). Where, “a” is the first term and “d” is the common difference among the series.
$
  {S_n}{\text{ = a + (a + d) + (a + 2d) + (a + 3d) + }}......{\text{ + [a + (n - 1)d]}} \\
  {{\text{S}}_{n{\text{ }}}} = \dfrac{n}{2}[2a + (n - 1)d] \\
$


Note: The geometric Progression is often called GP and arithmetic progression is called AP.
 For geometric Progression the formula changes depending on the ratio greater than or the less than one.
$ {S_n} = a(\dfrac{{1 - {r^n}}}{{1 - r}}),\;r < {\text{1}} $ Where, “a” is the first term and “r” is the ratio.