Question

Let \[{{S}_{n}}(1\le n\le 9)\] denotes the sum of \[n\] terms of series \[1+22+333+....+9999999999\], then for \[2\le n\le 9\]
A. \[{{S}_{n}}-{{S}_{n-1}}=\dfrac{1}{9}({{10}^{n}}-{{n}^{2}}+n)\]
B. \[{{S}_{n}}=\dfrac{1}{9}({{10}^{n}}-{{n}^{2}}+2n-2)\]
C. \[9\left( {{S}_{n}}-{{S}_{n-1}} \right)=n({{10}^{n}}-1)\]
D. None of these

Answer 1

Hint: In this question, we have to find the sum of \[n\] terms of the given series. For this, we need to split the series into a combination of series. Here we get all the series as geometric series. So, by using the sum of the \[n\] terms of a geometric series, we can find the required value.

Formula Used: If the series is a geometric series, then the sum of the $n$ terms is calculated by
 ${{S}_{n}}=\dfrac{a({{r}^{n}}-1)}{r-1}$ where $r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{n}}$ - Sum of the $n$ terms of the series; $n$ - Number of terms; $a$ - First term in the series; $r$ - is Common ratio.
The sum of $n$ natural numbers is $\dfrac{n(n+1)}{2}$.

Complete step by step solution: Given series is
\[{{S}_{n}}=1+22+333+....+9999999999\]
Rewriting the given series,
\[\begin{align}
  & {{S}_{n}}=\dfrac{1}{9}(9)+\dfrac{2}{9}(99)+\dfrac{3}{9}(999)+... \\
 & \text{ }=\dfrac{1}{9}\left( 10-1 \right)+\dfrac{2}{9}({{10}^{2}}-1)+\dfrac{3}{9}({{10}^{3}}-1)+... \\
 & \text{ }=\dfrac{1}{9}\left[ 10+{{2.10}^{2}}+{{3.10}^{3}}+... \right]-\dfrac{1}{9}\left[ 1+2+3+... \right] \\
\end{align}\]
We know that the sum of $n$ natural numbers is $\dfrac{n(n+1)}{2}$
Consider the obtained series as \[S=10+{{2.10}^{2}}+{{3.10}^{3}}+...+n\cdot {{10}^{n}}\]
Then, we get
\[{{S}_{n}}=\dfrac{1}{9}S-\dfrac{1}{9}\left[ \dfrac{n(n+1)}{2} \right]\text{ }...(1)\]
Now, simplifying the series \[S=10+{{2.10}^{2}}+{{3.10}^{3}}+...+n\cdot {{10}^{n}}\],
Multiplying by $10$ on both sides, we get
\[10S={{10}^{2}}+{{2.10}^{3}}+{{3.10}^{4}}+...(n-1){{10}^{n}}+n\cdot {{10}^{n+1}}\]
On subtracting the series $10S$ from the series $S$, we get
\[\begin{align}
  & S-10S=\left( 10+{{2.10}^{2}}+{{3.10}^{3}}+...+n\cdot {{10}^{n}} \right)-\left( {{10}^{2}}+{{2.10}^{3}}+{{3.10}^{4}}+...(n-1){{10}^{n}}+n\cdot {{10}^{n+1}} \right) \\
 & \Rightarrow -9S=\left( 10+{{10}^{2}}+{{10}^{3}}+...+\left( n\cdot {{10}^{n}}-(n-1){{10}^{n}} \right) \right)-n\cdot {{10}^{n+1}} \\
 & \Rightarrow -9S=\left( 10+{{10}^{2}}+{{10}^{3}}+...+{{10}^{n}} \right)-n\cdot {{10}^{n+1}} \\
\end{align}\]
\[\begin{align}
  & \Rightarrow 9S=n\cdot {{10}^{n+1}}-\left( 10+{{10}^{2}}+{{10}^{3}}+...+{{10}^{n}} \right) \\
 & \Rightarrow S=\dfrac{n}{9}{{10}^{n+1}}-\dfrac{1}{9}\left[ \dfrac{10({{10}^{n}}-1)}{10-1} \right] \\
 & \Rightarrow S=\dfrac{n}{9}{{10}^{n+1}}-\dfrac{10({{10}^{n}}-1)}{81} \\
 & \Rightarrow S=\dfrac{n}{9}{{10}^{n+1}}-\dfrac{{{10}^{n+1}}-10}{81}\text{ }...(2) \\
\end{align}\]
Then, substituting (2) in (1), we get
\[\begin{align}
  & {{S}_{n}}=\dfrac{1}{9}\left[ \dfrac{n}{9}{{10}^{n+1}}-\dfrac{{{10}^{n+1}}-10}{81} \right]-\dfrac{1}{9}\left[ \dfrac{n(n+1)}{2} \right] \\
 & \Rightarrow {{S}_{n}}=\dfrac{n}{81}{{10}^{n+1}}-\dfrac{{{10}^{n+1}}-10}{9\times 81}-\dfrac{1}{9}\left[ \dfrac{n(n+1)}{2} \right] \\
 & \Rightarrow 9{{S}_{n}}=\dfrac{9n\cdot {{10}^{n+1}}-{{10}^{n+1}}+10}{81}-\left[ \dfrac{n(n+1)}{2} \right] \\
 & \Rightarrow 9{{S}_{n}}=\dfrac{(9n-1){{10}^{n+1}}+10}{81}-\left[ \dfrac{n(n+1)}{2} \right] \\
 & \Rightarrow 9{{S}_{n}}=\dfrac{(9n-1){{10}^{n+1}}}{81}+\dfrac{10}{81}-\left[ \dfrac{n(n+1)}{2} \right]\text{ }...(3) \\
\end{align}\]
Now, substituting $n=n-1$ in (3) we get,
\[\begin{align}
  & 9{{S}_{n-1}}=\dfrac{(9(n-1)-1){{10}^{(n-1)+1}}}{81}+\dfrac{10}{81}-\left[ \dfrac{(n-1)((n-1)+1)}{2} \right] \\
 & \text{ }=\dfrac{(9n-9-1){{10}^{n}}}{81}+\dfrac{10}{81}-\left[ \dfrac{n(n-1)}{2} \right] \\
 & \text{ }=\dfrac{(9n-10){{10}^{n}}}{81}+\dfrac{10}{81}-\left[ \dfrac{n(n-1)}{2} \right]\text{ }...(4) \\
\end{align}\]
Then, by subtracting (4) from (3), we get
\[\begin{align}
  & 9{{S}_{n}}-9{{S}_{n-1}}=\dfrac{(9n-1){{10}^{n+1}}}{81}+\dfrac{10}{81}-\left[ \dfrac{n(n+1)}{2} \right]-\dfrac{(9n-10){{10}^{n}}}{81}-\dfrac{10}{81}+\left[ \dfrac{n(n-1)}{2} \right] \\
 & \text{ }=\dfrac{(9n-1){{10}^{n+1}}}{81}-\dfrac{(9n-10){{10}^{n}}}{81}+\left[ \dfrac{n(n-1)}{2} \right]-\left[ \dfrac{n(n+1)}{2} \right] \\
 & \text{ }=\dfrac{{{10}^{n}}}{81}\left[ (9n-1)10-(9n-10) \right]+\dfrac{n}{2}\left[ n-1-n-1 \right] \\
\end{align}\]
\[\begin{align}
  & \Rightarrow 9{{S}_{n}}-9{{S}_{n-1}}=\dfrac{{{10}^{n}}}{81}\left[ 90n-10-9n+10 \right]-n \\
 & \text{ }=\dfrac{{{10}^{n}}}{81}(81n)-n \\
 & \text{ }=n({{10}^{n}}-1) \\
\end{align}\]
Therefore, the required answer is \[9\left( {{S}_{n}}-{{S}_{n-1}} \right)=n({{10}^{n}}-1)\].

Option ‘C’ is correct

Note: Here we need to split the given series in order to avoid the complexity of the problem in solving. Since all the series here are geometric series. So, we can easily calculate their sums. By using their sums, we can extract the required expression.