Question

Find the sum to n terms of the sequence 8,88,888,8888... \[\]

Answer 1

Hint: We denote the sum of 8,88,888,8888... up to $n$ terms as $S=8+88+888...$. We first take 8 common and then multiply and divide by 9 the right hand side of the equation. We express the terms 9, 99, 999 ... in terms of the nearest multiple of 10 and where we find two series ${{S}_{1}}=10+{{10}^{2}}+{{10}^{3}}...\left( \text{upto }{{n}^{\text{th}}}\text{ term} \right)$ and${{S}_{2}}=1+1+1...\left( \text{upto }{{n}^{\text{th}}}\text{ term} \right)$. We find ${{S}_{1}}$ with GP series in formula for first $n$ terms $\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ and then ${{S}_{2}}$.\[\]

Complete step by step answer:
A sequence is defined as the enumerated collection of numbers where repetitions are allowed and order of the numbers matters. The members of the sequence are called terms. Mathematically, a sequence with infinite terms is written as
\[\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...\]
If we add up the terms then the sequence is called series. The series for above sequence is
\[{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...\]
Geometric sequence otherwise known as Geometric progression, abbreviated as GP is a type sequence where the ratio between any two consecutive numbers is constant. If $\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...$ is an GP, then
\[\dfrac{{{x}_{2}}}{{{x}_{1}}}=\dfrac{{{x}_{3}}}{{{x}_{2}}}...=r\]
Here the ratio between two terms is called the common ratio and denoted as $r$. The first term is conventionally denoted as $a$. Then the GP series for $n$ terms is given by
\[a+ar+a{{r}^{2}}+...a{{r}^{n-1}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}\]
We are given the sequence in the question as
\[8,88,888,8888,....\]
Let us denote the sum of first $n$ terms of the sequence as $S.$ We have,
\[S=8+88+888+...\left( \text{upto }{{n}^{\text{th}}}\text{ terms} \right)\]
Let us take 8 common from the right hand side of the above equation and get,
\[\Rightarrow S=8\left( 1+11+111...\text{upto }{{n}^{\text{th}}}\text{ term} \right)\]
Let us divide and multiply by 9 in the right hand side of the above equation and get,
\[\Rightarrow S=\dfrac{9}{9}\times 8\left( 1+11+111...\text{upto }{{n}^{\text{th}}}\text{ term} \right)\]
Let us take 9 in the numerator inside the bracket and get,
\[\begin{align}
  & \Rightarrow S=\dfrac{1}{9}\times 8\left( 9+99+999...\text{upto }{{n}^{\text{th}}}\text{ term} \right) \\
 & \Rightarrow S=\dfrac{8}{9}\left( 9+99+999...\text{upto }{{n}^{\text{th}}}\text{ term} \right) \\
\end{align}\]
Let us write the 9, 99,999... in terms of nearest multiple of 10. We have,
\[\begin{align}
  & \Rightarrow S=\dfrac{8}{9}\left( 10-1+100-1+1000-1+...\text{upto }{{n}^{\text{th}}}\text{ term} \right) \\
 & \Rightarrow S=\dfrac{8}{9}\left\{ \left( 10+100+100+...\text{upto }{{n}^{\text{th}}}\text{ term} \right)-\left( 1+1+1...\text{upto }{{n}^{\text{th}}}\text{ term} \right) \right\} \\
 & \Rightarrow S=\dfrac{8}{9}\left\{ \left( 10+{{10}^{2}}+{{10}^{3}}+...\text{upto }{{n}^{\text{th}}}\text{ term} \right)-\left( 1+1+1...\text{upto }{{n}^{\text{th}}}\text{ term} \right) \right\} \\
 & \Rightarrow S=\dfrac{8}{9}\left\{ {{S}_{1}}-{{S}_{2}} \right\}....\left( 1 \right) \\
\end{align}\]
Let us find the sum ${{S}_{1}}=10+{{10}^{2}}+{{10}^{3}}+...\text{upto }{{n}^{\text{th}}}\text{ term}$ first. We see that ${{S}_{1}}$ is a GP series with first term $a=10$ and common ratio $r=\dfrac{{{10}^{2}}}{10}=\dfrac{{{10}^{3}}}{{{10}^{2}}}=10$. So we use the formula for sum of first $n$ terms in a GP sequence and get ${{S}_{1}}$ as
\[ {{S}_{1}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}=\dfrac{10\left( {{10}^{n}}-1 \right)}{10-1}=\dfrac{10}{9}\left( {{10}^{n}}-1 \right)\]

We now find the sum ${{S}_{2}}=1+1+1...\text{upto }{{n}^{\text{th}}}\text{ term}$. If we add 1 for $n$ times, the sum will be $1\times n$ . So we have
\[{{S}_{2}}=1\times n=n\]
Let us put ${{S}_{1}},{{S}_{2}}$ in equation (1) and have,
\[\begin{align}
  & \Rightarrow S=\dfrac{8}{9}\left( \dfrac{10}{9}\left( {{10}^{n}}-1 \right)-n \right) \\
 & \Rightarrow S=\dfrac{80}{81}\left( {{10}^{n}}-1 \right)-\dfrac{8}{9}n \\
\end{align}\]

Note: We can find the GP series with infinite terms if $\left| r \right|<1$ as $S=\dfrac{a}{1-r}$. The ${{n}^{t\text{h}}}$ term of the given sequence 8, 88,888 ... is $\dfrac{8}{9}\left( {{10}^{n}}-1 \right)$. We must be careful of the confusion between the formula for GP series from AP series which is given by ${{S}_{n}}=\dfrac{n}{2}\left( a+\left( n-1 \right)d \right)$ where $d$ is the common difference.