Question

If $\left| x \right|<1$, then the sum of the series $1+2x+3{{x}^{2}}+4{{x}^{3}}+.....\infty $ will be
A. $\dfrac{1}{1-x}$
B. $\dfrac{1}{1+x}$
C. $-\dfrac{1}{1+{{x}^{2}}}$
D. \[\dfrac{1}{{{\left( 1-x \right)}^{2}}}\]

Answer 1

Hint: We use \[{{\left( 1-x \right)}^{n}}=1-nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}-......+{{\left( -1 \right)}^{r}}\dfrac{n\left( n-1 \right)....\left( n-r+1 \right)}{r!}{{x}^{r}}+....+\infty \], the infinite binomial formula. We replace the value of $n=-2$. We simplify the coefficients and get the simplest forms. We convert the indices to get the required solution.

Complete answer:
We know that the binomial theorem of \[{{\left( 1+x \right)}^{n}},\left| x \right|<1,n\in \mathbb{R}\] gives
\[{{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}+......+\dfrac{n\left( n-1 \right)....\left( n-r+1 \right)}{r!}{{x}^{r}}+....+\infty \].
Now when the value of $n$ is a negative real number we get
\[{{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}+......+\dfrac{n\left( n-1 \right)....\left( n-r+1 \right)}{r!}{{x}^{r}}+....+\infty \]
We first replace the value of $x$ with $-x$ to get
\[{{\left( 1-x \right)}^{n}}=1-nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}-......+{{\left( -1 \right)}^{r}}\dfrac{n\left( n-1 \right)....\left( n-r+1 \right)}{r!}{{x}^{r}}+....+\infty \].
We now put the value for $n=-2$ to get
\[{{\left( 1-x \right)}^{-2}}=1-\left( -2 \right)x+\dfrac{\left( -2 \right)\left( -2-1 \right)}{2!}{{x}^{2}}-\dfrac{\left( -2 \right)\left( -2-1 \right)\left( -2-2 \right)}{3!}{{x}^{3}}+.......\infty \].
We now simplify to get
\[\begin{align}
  & {{\left( 1-x \right)}^{-2}} \\
 & =1-\left( -2 \right)x+\dfrac{\left( -2 \right)\left( -2-1 \right)}{2!}{{x}^{2}}-\dfrac{\left( -2 \right)\left( -2-1 \right)\left( -2-2 \right)}{3!}{{x}^{3}}+.......\infty \\
 & =1+2x+3{{x}^{2}}+4{{x}^{3}}+.....\infty \\
\end{align}\]
Therefore, the required sum is \[{{\left( 1-x \right)}^{-2}}=\dfrac{1}{{{\left( 1-x \right)}^{2}}}\].
And hence the correct answer is option D.

Note:
An infinite series has a sum that is convergent in the limit sense. The limiting value can be found from \[{{\left( 1+x \right)}^{n}}\]. In the case of a fraction value of $n$ we can also use the infinite form. It can take both negative and positive but for an integer, it has to be negative.