Question

Find the sum to n terms of the sequence 8, 88, 888, 8888, ....?

Answer 1

This question is based on the topic A.G.P(Arithmetic Geometric progression). Firstly write the given expression in summation form up to ‘n’ terms and let it be S. Again write it but shift the numbers in RHS by one place. After this subtract the both equations to get the nth term. Use the nth term expression to find the summation.

Complete step-by-step answer:
In the question we have to find the sum (${{\text{S}}_n}$) = 8+88+888+.... up to n terms.
Let the required sum be ${{\text{S}}_n}$ .
$$\therefore$$${{\text{S}}n}$= 8+88+888+... +${{\text{T}}{{\text{n - 1}}}} + {{\text{T}}{\text{n}}}$------------(1) Writing the above summation again but shifting RHS by one terms, we get:${{\text{S}}n}$= 0+ 8+88+888+... +${{\text{T}}{{\text{n - 1}}}} + {{\text{T}}{\text{n}}}$--------------(2) On subtracting the equation 2 from equation 1, we get: 0 = 8+80+800+8000+.........+${\text{(}}{{\text{T}}{\text{n}}} - {{\text{T}}{{\text{n - 1}}}}{\text{) - }}{{\text{T}}{\text{n}}}$$ \Rightarrow {{\text{T}}{\text{n}}}$= 8+80+800+8000+......+${\text{(}}{{\text{T}}{\text{n}}} - {{\text{T}}{{\text{n - 1}}}}{\text{)}}$-------- (3) In the RHS, there are total ‘n’ terms. The terms in equation 3 are in G.P. We know that the sum of ‘n’ terms of a G.P. is given by:${\text{Sum = }}\dfrac{{{\text{a}}\left( {{{\text{r}}^n} - 1} \right)}}{{{\text{r - 1}}}};{\text{r > 1}}$ ------------ (4)
Where, ‘a’ is the first term of G.P.
‘r’ is the common ratio.
And ‘n’ is the number of terms.

From the G.P. series given by equation 3, we can say that:
a = 8
r = 10
And number of terms = n
Using equation 3 and equation 4, we can write:
${{\text{T}}_{\text{n}}} = \dfrac{{8\left( {{{10}^n} - 1} \right)}}{{{\text{10 - 1}}}} = \dfrac{{8\left( {{{10}^n} - 1} \right)}}{9}$ .
We know that sum is also given as:
${{\text{S}}_n} = \sum\limits_{n = 1}^n {{{\text{T}}_{\text{n}}}}$ ---------- (5)
Putting the value of ${{\text{T}}_{\text{n}}}$ in equation 5, we get:
${{\text{S}}_n} = \sum\limits_{n = 1}^n {\dfrac{{8\left( {{{10}^n} - 1} \right)}}{9} = } \dfrac{8}{9}\sum\limits_{n = 1}^n {\left( {{{10}^n} - 1} \right)} = \dfrac{8}{9}\left( {\sum\limits_{n = 1}^n {{{10}^n} - \sum\limits_{n = 1}^n 1 } } \right)$ ------------ (6)
Now, the first summation can written as;
$\sum\limits_{n = 1}^n {{{10}^{\text{n}}}} = 10 + {10^2} + {10^3} + ..... + {10^{\text{n}}}$ .
This is a G.P.. So using the formula for finding G.P., we get:
$\sum\limits_{n = 1}^n {{{10}^{\text{n}}}} = \dfrac{{10\left( {{{10}^n} - 1} \right)}}{{{\text{10 - 1}}}} = \dfrac{{10\left( {{{10}^n} - 1} \right)}}{9}$ ----------(7)
Also, the second summation can written as;
$\sum\limits_{n = 1}^n 1$ =1+1++......+ n times = n . --------- (8)

Putting the value of equation 7 and 8 in equation 6, we get:
${{\text{S}}_n} = \dfrac{8}{9}\left( {\dfrac{{10\left( {{{10}^n} - 1} \right)}}{9} - n} \right)$ .

Note: In series summation type of question, you have to first recognise the type of progression. In this case, it is A.G.P(Arithmetic Geometric Progression). So you have to follow the procedure which we have used in this case. You should know the basic formula ${{\text{S}}_n} = \sum\limits_{n = 1}^n {{{\text{T}}_{\text{n}}}}$ to find summation once you get the nth term. You should remember the formula for finding the sum of terms of a G.P.