Question

Find the sum to n terms:
 $ 2 + 22 + 222 + .........\,\,upto\;\,n\,\,terms $

Answer 1

Hint: For this type of series in we first take number which is repeating in each term common and then multiply and divide the result by number $ 9 $ and then writing all different terms having $ 9,\,\,99,\,\,999... $ in term of $ 10,100,1000..... $ and then using formula for sum of n terms in G.P. to find required solution of the problem.
Sum of n terms in G.P. $ a\left( {\dfrac{{{r^n} - 1}}{{r - 1}}} \right),r > 1\,\,\,and\,\,\,a\left( {\dfrac{{1 - {r^n}}}{{1 - r}}} \right),\,\,r < 1 $

Complete step-by-step answer:
Given,
 $ 2 + 22 + 222 + .........\,\,upto\;\,n\,\,terms $
Taking $ 2 $ common from above. We have,
 $ 2\left( {1 + 11 + 111 + ...................\,\,upto\,\,n\,terms} \right) $
Multiplying and divide by $ 9 $
 $ \dfrac{2}{9}\left( {9 + 99 + 999 + ...................upto\,\,n\,\,terms} \right) $
Now, writing $ 9\,\,as\,\,\,\left( {10 - 1} \right),\,\,99\,\,as\,\,\left( {100 - 1} \right)\,\,and\,\,999\,\,as\,\,\left( {1000 - 1} \right)\,\,and\,\,\,so\,\,on $ in above series. We have
 $ \dfrac{2}{9}\left\{ {\left( {10 - 1} \right) + \left( {100 - 1} \right) + \left( {1000 - 1} \right) + ..........upto\,\,n\,\,terms} \right\} $
Separating terms and writing it as:
 $ \dfrac{2}{9}\left\{ {\left( {10 + 100 + 1000 + ...........upto\,\,n\,\,terms} \right) - 1 - 1 - 1 - 1 - 1........upto\,\,n\,\,terms} \right\} $
Or we can write it as:
 $ \dfrac{2}{9}\left\{ {\left( {10 + {{10}^2} + {{10}^3} + ........up\,to\,n\,\,terms} \right) - n} \right\} $
Clearly the first series in above equations is a G.P. therefore, using the sum of n terms in G.P. We have,
 $ \dfrac{2}{9}\left\{ {10\left( {\dfrac{{{{10}^n} - 1}}{{10 - 1}}} \right) - n} \right\}\left\{ {\because {S_n} = a\left( {\dfrac{{{r^n} - 1}}{{r - 1}}} \right)} \right\} $
 $
   \Rightarrow \dfrac{{20}}{9}\left( {\dfrac{{{{10}^n} - 1}}{9}} \right) - \dfrac{2}{9}n \\
   \Rightarrow \dfrac{{20}}{{81}}\left( {{{10}^n} - 1} \right) - \dfrac{2}{9}n \;
  $
Which is the required sum of the series.
Hence, from above we see that sum of the series $ 2 + 22 + 222 + ......... $ is $ \dfrac{{20}}{{81}}\left( {{{10}^n} - 1} \right) - \dfrac{2}{9}n $ .

Note: In geometric progression to find sum of n terms there are two formulas. There is a one formula for those series in which the common ratio is less than one and another is for in which the common ratio is greater than one. So, while doing this type of problem one must choose a formula according to the value of ‘r’ to get the correct solution of the problem.