Question

Find the sum of the sequence $3,33,333,3333, \ldots \ldots $to $n$ terms.

Answer 1

Hint: In this problem, to find the sum of the given sequence first we will write $3 + 33 + 333 + 3333 + \ldots \ldots $. Consider up to $n$ terms. We will simplify this series to get the required sum. First we will multiply and divide by $3$. After simplification we will get terms in geometric progression. So, we will use the formula of the sum of first $n$ terms of geometric series.

Complete step-by-step answer:
Let us denote the sum by $S$. Therefore, $S = 3 + 33 + 333 + 3333 + \ldots \ldots $($n$ terms)
Let us multiply and divide by $3$. Therefore, $S = \dfrac{1}{3}\left[ {9 + 99 + 999 + 9999 + \ldots \ldots } \right]$
Now we will simplify this sum. Therefore,
$
  S = \dfrac{1}{3}\left[ {\left( {10 - 1} \right) + \left( {100 - 1} \right) + \left( {1000 - 1} \right) + \ldots \ldots } \right] \\
   \Rightarrow S = \dfrac{1}{3}\left[ {\left( {10 + 100 + 1000 + \ldots \ldots } \right) - \left( {1 + 1 + 1 + \ldots \ldots } \right)} \right] \\
   \Rightarrow S = \dfrac{1}{3}\left[ {\left( {10 + 100 + 1000 + \ldots \ldots } \right) - n} \right] \cdots \cdots \left( 1 \right) \\
$
Note that here we need to find the sum of n terms. So, $1 + 1 + 1 + \ldots \ldots $($n$ terms) $ = n$
Now we can say that $10 + 100 + 1000 + \ldots \ldots $ is a geometric series because the ratio of two consecutive terms are equal. That is, the ratio is $10$.
Now we are going to use the formula of the sum of first $n$ terms geometric series. Sum of first $n$ terms of geometric series is $\dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$ where $a$ is the first term of the series and $r$ is the common ratio of two consecutive terms.
In the series $10 + 100 + 1000 + \ldots \ldots $, the first term is $a = 10$ and ratio $r = 10.$
Now, from equation $\left( 1 \right)$, we get
$
  S = \dfrac{1}{3}\left[ {\dfrac{{10\left( {{{10}^n} - 1} \right)}}{{10 - 1}} - n} \right] \\
   \Rightarrow S = \dfrac{1}{3}\left[ {\dfrac{{10\left( {{{10}^n} - 1} \right)}}{9} - n} \right] \\
   \Rightarrow S = \dfrac{1}{{27}}\left[ {10\left( {{{10}^n} - 1} \right) - 9n} \right] \\
$
Hence, the required sum is $\dfrac{1}{{27}}\left[ {10\left( {{{10}^n} - 1} \right) - 9n} \right]$.

Note: A sequence is defined as an arrangement of numbers (elements) in a fixed order. The sum of the elements of the sequence is called series. In the given problem, we have to consider only $n$ terms. So, it is called the finite series. The sum of infinite numbers of the sequence is called an infinite series. If the difference between two consecutive terms of the sequence is equal (constant) then it is called arithmetic sequence. If the ratio of two consecutive terms of the sequence is equal (constant) then it is called geometric sequence. We can find the sum of first $n$ terms of arithmetic sequence as well as geometric sequence by using formulas without adding those $n$ terms. The series $a + ar + a{r^2} + a{r^3} + \ldots \ldots + a{r^{n - 1}} + ......$ is called geometric series and the sum of first $n$ terms of this series is given by $\dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$. This sum is valid only if $r \ne 1$ because if $r = 1$ then the denominator becomes zero.