Question

Find the sum of the series
$1\times 1+3\times 0.1+5\times 0.01+7\times 0.001+.........n\, terms$

Answer 1

Hint: Here we have to find the sum of the series which are neither in A.P nor in G.P. We will convert the following series in G.P. For that, we will first multiply all the terms of the series with $0.1$ and then we will subtract the new series with the original series. The series which we will get after subtraction will be in G.P. we can easily find its sum using the formula of sum of G.P.

Complete step-by-step answer:
Let the sum of the series $1\times 1+3\times 0.1+5\times 0.01+7\times 0.001+.........n\, terms $be${{S}_{n}}$.
$1\times 1+3\times 0.1+5\times 0.01+7\times 0.001+.........n\, terms={{S}_{n}}..........\left( 1 \right)$
Now, we will multiply on both sides.
$\Rightarrow 0.1\left( 1\times 1+3\times 0.1+5\times 0.01+7\times 0.001+.........n\, terms \right)=0.1{{S}_{n}}$
Multiplying 0.1 with each term, we get
$\Rightarrow 1\times 0.1+3\times 0.01+5\times 0.001+7\times 0.0001+.........n\,terms=0.1{{S}_{n}}..........\left( 2 \right)$
We will subtract equation 2 from equation 1 now.
$\begin{align}
  & \underline{\begin{align}
  & 1\times 1+3\times 0.1+5\times 0.01+7\times 0.001+.........n\, terms={{S}_{n}} \\
 & -0\times 1-1\times 0.1-3\times 0.01-5\times 0.001-.........n \,terms=-0.1{{S}_{n}} \\
\end{align}} \\
 & 1\times 1+2\times 0.1+2\times 0.01+2\times 0.001+.........n\, terms=0.9{{S}_{n}} \\
\end{align}$
We can write the resulting series as
$1+2\left( 0.1+0.01+0.001+........n\, terms \right)=0.9{{S}_{n}}..........\left( 3 \right)$
The series inside the bracket is in G.P as the common ratio between the consecutive terms is $\dfrac{0.01}{0.1}=0.1$
Therefore, sum of the series $\left( 0.1+0.01+0.001+........n\, terms \right)=0.1\dfrac{1-{{\left( 0.1 \right)}^{n}}}{1-0.1}=\dfrac{1-{{\left( 0.1 \right)}^{n}}}{9}$
We will put the value of the sum of series in equation 3.
$\Rightarrow 1+2\dfrac{1-{{\left( 0.1 \right)}^{n}}}{9}=0.9{{S}_{n}}$
On simplifying the terms, we get
$\Rightarrow \dfrac{9+2-2{{\left( 0.1 \right)}^{n}}}{9}=0.9{{S}_{n}}$
On further simplification, we get
$\Rightarrow {{S}_{n}}=\dfrac{11+2{{\left( 0.1 \right)}^{n}}}{8.1}$
Thus, the sum of the series $1\times 1+3\times 0.1+5\times 0.01+7\times 0.001+.........n\, terms$ is $\dfrac{11+2{{\left( 0.1 \right)}^{n}}}{8.1}$.

Note: Here we have changed the following series into G.P which means Geometric Progression. A sequence of series is said to be a geometric progression if the ratio of consecutive terms of the series is always a constant.
Let the series $a+b+c+d+.......$ is a geometric progression, then the ratio of the consecutive terms is constant and equal.
$\dfrac{b}{a}=\dfrac{c}{b}=k$
Here $k$ is constant.
Some important properties of the G.P are:-
If we multiply or divide a non zero number to each term of the G.P, then the resulting sequence is also in G.P and the common ratio remains the same.
The reciprocal of all the terms of the geometric progression will have their common ratios constant i.e. the resulting sequence will also be in geometric progression.