Question

Evaluate ${\text{7 + 77 + 777 + }}........{\text{ + }}$up to n terms
${\text{(A) }}\dfrac{7}{{81}}[{10^{n + 1}} - 9n]$
${\text{(B) }}\dfrac{7}{{81}}[{10^{n + 1}} - 9n - 10]$
${\text{(C) }}\dfrac{7}{{81}}[{10^n} - 9n - 10]$
${\text{(D) }}\dfrac{7}{{81}}[{10^{n + 1}} - n - 10]$

Answer 1

In this question we will first modify the given expression into the form of a geometric progression so that it could be represented by a formula. Then we will get the required answer.
We will make use of the sum of n terms formula which is given by
${S_n} = \dfrac{{a({r^n} - 1)}}{{r - 1}}$
Here $a$ is the first term in the series, and $r$ is the division of consecutive terms and $n$ is the number of terms in the sequence.

Complete step-by-step answer:
It is given that the question stated as, ${\text{7 + 77 + 777 + }}........{\text{ + upto n terms}}$
Since $7$ is a common divisible for all the numbers, we take it as common,
$= 7(1{\text{ + 11 + 111 + }}........{\text{ + upto n terms)}}$
Now we will multiply and divide the expression by $9$
$= \dfrac{7}{9}(9 \times 1{\text{ + 9}} \times {\text{11 + 9}} \times {\text{111 + }}........{\text{ + upto n terms)}}$
On simplifying the expression, we get:
$= \dfrac{7}{9}(9{\text{ + 99 + 999 + }}........{\text{ + upto n terms)}}$
Since $9$ can be written as$10 - 1$, $99$ can be written as $100 - 1$ and so on, we make this change in the expression.
$= \dfrac{7}{9}((10 - 1){\text{ + (100}} - {\text{1) + (1000}} - {\text{1) + }}........{\text{ + upto n terms)}}$
On opening the bracket, we get:
$= \dfrac{7}{9}(10 - 1{\text{ + 100}} - 1{\text{ + 1000}} - {\text{1 + }}........{\text{ + upto n terms)}}$
Since there are $n$ terms in the distribution so there the total number of $- 1$ in the expression will be the same as there are terms in the expression
$= \dfrac{7}{9}(10{\text{ + 100 + 1000 + }}........{\text{ + upto n terms}} - n{\text{)}}$
Now $10$ can be written as ${10^1}$, $100$ can be written as ${10^2}$ and so on, therefore the expression can be written as:
$= \dfrac{7}{9}({10^1}{\text{ + 1}}{{\text{0}}^2}{\text{ + 1}}{{\text{0}}^3}{\text{ + }}........{\text{ + upto n terms}} - n{\text{)}}$
Since all the terms are in a geometric progression of $n$ terms where $a = 10$ and $r = \dfrac{{{{10}^2}}}{{10}} = 10$
Now we use the formula, $Sn = \dfrac{{a({r^n} - 1)}}{{r - 1}}$
On substituting the values, we get:
$Sn = \dfrac{{10({{10}^n} - 1)}}{{10 - 1}}$
On simplifying we get:
$Sn = \dfrac{{10({{10}^n} - 1)}}{9}$
Now we will add this formula to the expression:
$= \dfrac{7}{9}\left( {\dfrac{{10({{10}^n} - 1)}}{9} - n} \right)$
On multiply the bracket terms and we get:
$= \dfrac{7}{9}\left( {\dfrac{{{{10}^{n + 1}} - 10}}{9} - n} \right)$
Taking LCM we get,
$= \dfrac{7}{9}\left( {\dfrac{{{{10}^{n + 1}} - 10 - 9n}}{9}} \right)$
On multiply the denominator term we get,
$= \dfrac{7}{{81}}(({10^{n + 1}} - 10) - 9n{\text{)}}$
It could be further simplified as:
$= \dfrac{7}{{81}}({10^{n + 1}} - 9n - 10{\text{)}}$

Therefore, the correct answer is option $(B)$.

Note: In these types of questions try to convert the expressions in a format such that it could be represented using a formula of Arithmetic progression or geometric progression.
The general formula of an arithmetic progression is $a$,$a + d$,$a + 2d$, …
The ${n^{th}}$ term of the arithmetic progression is ${{\text{T}}_{\text{n}}}{\text{ = a + (n - 1)d}}$
Here, $d =$ common difference, $a =$ first term, ${{\text{T}}_{\text{n}}}{\text{ = }}{{\text{n}}^{{\text{th}}}}{\text{ term}}$
If we selected terms will be in Arithmetic progression, then the term in the regular interval form arithmetic.