Question

If the sum of the series $6+66+666+....n\text{ terms}$ is
A. $[\dfrac{({{10}^{n-1}}-9n+10)}{81}]$
B. $[\dfrac{2({{10}^{n+1}}-9n-10)}{27}]$
C. $[\dfrac{2({{10}^{n}}-9n+10)}{27}]$
D. None of these

Answer 1

Hint: In this question, we need to find the sum of the n terms given in the series. By applying basic algebraic operations, the series becomes a geometrical progression or geometric series. Then, by applying appropriate formulae the sum is calculated.

Formula used:The geometric sequence is written as
$a,ar,a{{r}^{2}},a{{r}^{3}},...,a{{r}^{n}}$; $a$ is the first term and $r$ is the common ratio of the series.
The common ratio is calculated by
$r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
The sum of the n terms is calculated by
 ${{S}_{n}}=\dfrac{a({{r}^{n}}-1)}{r-1}$


Complete step by step solution: The given series is $6+66+666+....n\text{ terms}$
On applying basic algebraic operations to the given series, we get
$\begin{align}
  & 6+66+666+....n\text{ terms} \\
 & \text{=6(1+11+111+}....n\text{ terms}) \\
\end{align}$
On multiplying and dividing by 9, we get
$\begin{align}
  & 6+66+666+....n\text{ terms} \\
 & \text{=}\dfrac{6}{9}\text{(9+99+999+}....n\text{ terms}) \\
\end{align}$
On simplifying,
$[\begin{align}
  & \text{=}\dfrac{6}{9}\left( (10-1)+(100-1)+(1000-1)+....n\text{ terms} \right) \\
 & =\dfrac{6}{9}\left( (10-1)+({{10}^{2}}-1)+({{10}^{3}}-1)+...n\text{ terms} \right) \\
 & =\dfrac{6}{9}\left[ (10+{{10}^{2}}+{{10}^{3}}+...n\text{ terms})-(1+1+1+...n\text{ times)} \right] \\
 & =\dfrac{6}{9}\left[ (10+{{10}^{2}}+{{10}^{3}}+...n\text{ terms})-n \right]\text{ }...\text{(1)} \\
\end{align}]$
In the above equation (1), the series formed is a geometric series.
So,
$\begin{align}
  & a=10; \\
 & r=\dfrac{{{10}^{2}}}{10}=10 \\
\end{align}$Then, the sum of the n terms in the obtained series is
$\begin{align}
  & {{S}_{n}}=\dfrac{a({{r}^{n}}-1)}{r-1} \\
 & \text{ }=\dfrac{10({{10}^{n}}-1)}{10-1} \\
 & \text{ }=\dfrac{10}{9}({{10}^{n}}-1) \\
\end{align}$
On substituting the obtained sum in equation (1), we get
$[\begin{align}
  & =\dfrac{6}{9}\left[ (\dfrac{10({{10}^{n}}-1)}{9})-n \right] \\
 & =\dfrac{6}{9}\left[ \dfrac{10({{10}^{n}}-1)-9n}{9} \right] \\
 & =\dfrac{6}{81}\left[ 10({{10}^{n}}-1)-9n \right] \\
 & =\dfrac{2}{27}\left[ {{10}^{n+1}}-9n-10 \right] \\
\end{align}]$


Thus, Option (B) is correct.

Note: Here the given series is a geometric series. By using the appropriate formula for finding the sum of n terms, the required value is calculated. Here we may go wrong with the formula for finding the sum of n terms and the sum of infinite terms of a geometric series. In this question, we get an infinite series, on rewriting the given series. So, the proper formula is applied to get the required sum.