Question

The sum of n terms of the following series $1+(1+x)+(1+x+{{x}^{2}})+....n\text{ terms}$ will be
A. \[\dfrac{1-{{x}^{n}}}{1-x}\]
B. \[\dfrac{x(1-{{x}^{n}})}{1-x}\]
C. \[\dfrac{n(1-x)-x(1-{{x}^{n}})}{{{(1-x)}^{2}}}\]
D. None of these

Answer 1

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

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

Complete step by step solution: The given series is $1+(1+x)+(1+x+{{x}^{2}})+....n\text{ terms}$
On multiplying and dividing by $(1-x)$, we get
$\begin{align}
 & =\dfrac{1}{(1-x)}\left[ (1-x)+(1-x)(1+x)+(1-x)(1+x+{{x}^{2}})+....n\text{ terms} \right] \\
 & =\dfrac{1}{(1-x)}\left[ (1-x)+(1-{{x}^{2}})+(1-{{x}^{3}})+....n\text{ terms} \right] \\
\end{align}$
On separating the common terms,
\[\begin{align}
  & =\dfrac{1}{(1-x)}\left[ (1+1+1+...n\text{ times})-(x+{{x}^{2}}+{{x}^{3}}+...n\text{ terms}) \right] \\
 & =\dfrac{1}{(1-x)}\left[ n-(x+{{x}^{2}}+{{x}^{3}}+...n\text{ terms}) \right]\text{ }...\text{(1)} \\
\end{align}\]
In the above equation (1), the series formed is a geometric series.
So,
$\begin{align}
  & a=x; \\
 & r=\dfrac{{{x}^{2}}}{x}=x \\
\end{align}$
Then, the sum of the n terms in the obtained series is
$\begin{align}
  & {{S}_{n}}=\dfrac{a(1-{{r}^{n}})}{1-r} \\
 & \text{ }=\dfrac{x(1-{{x}^{n}})}{1-x} \\
\end{align}$
Substituting the obtained sum in equation (1), we get
\[\begin{align}
  & =\dfrac{1}{(1-x)}\left[ n-(\dfrac{x(1-{{x}^{n}})}{1-x}) \right] \\
 & =\dfrac{1}{(1-x)}\left[ \dfrac{n(1-x)-x(1-{{x}^{n}})}{1-x} \right] \\
 & =\dfrac{1}{(1-x)}\left[ \dfrac{n(1-x)-x(1-{{x}^{n}})}{1-x} \right] \\
 & =\dfrac{n(1-x)-x(1-{{x}^{n}})}{{{(1-x)}^{2}}} \\
\end{align}\]

Thus, Option (C) is correct.

Note: Here the given series forms a geometric series. By using the appropriate formula for finding the sum of n terms, the required value is calculated. Avoid calculation errors.