Question

# The sum to infinity of the series $1 - 3x + 5{x^2} - 7{x^3} + ....\infty$, when $\left| x \right| < 1$, is A.${S_\infty } = \dfrac{{1 - x}}{{{{\left( {1 + x} \right)}^2}}}$B.${S_\infty } = \dfrac{{1 - x}}{{{{\left( {1 + x} \right)}^3}}}$C.${S_\infty } = \dfrac{{1 + x}}{{{{\left( {1 - x} \right)}^2}}}$D.${S_\infty } = \dfrac{{1 + x}}{{{{\left( {1 - x} \right)}^3}}}$

Hint: We will first form the G.P. using the given series. We will first multiply the given series by $x$ and then we will then subtract the two equations to form a G.P. We will then calculate the sum using the formula ${S_\infty } = \dfrac{a}{{1 - r}}$, where $a$ is the first term and $r$ is the common ratio and $\left| r \right| < 1$

We are given that the series is $S = 1 - 3x + 5{x^2} - 7{x^3} + ....\infty$ (1)
Now, we will multiply the equation (1) by $x$
Then, we will get $Sx = x - 3{x^2} + 5{x^3} - 7{x^4} + ....\infty$ (2)
Now, subtract equation (1) from equation (2)
That is,
$S - Sx = 1 - 3x + 5{x^2} - 7{x^3} + 3x - 5{x^2} + ...\infty$
Which is equals to
$S\left( {1 - x} \right) = 1 + 2x + 2{x^2} + 2{x^3}...\infty$
We can write the above equation as
$S\left( {1 - x} \right) = 1 + 2\left( {x + {x^2} + {x^3}...\infty } \right)$
Now, we can see the series $x + {x^2} + {x^3}...\infty$ is a G.P. with a common ratio as $x$.
As, we know the sum of infinite number of terms of a G.P. is given by
${S_\infty } = \dfrac{a}{{1 - r}}$, where $\left| r \right| < 1$
We have $a = x$, $r = x$ and $\left| x \right| < 1$
Then, the sum of series $x + {x^2} + {x^3}...\infty$ is $\dfrac{x}{{1 - x}}$
Therefore, on substituting the value of $x + {x^2} + {x^3}...\infty$ in equation $S\left( {1 - x} \right) = 1 + 2\left( {x + {x^2} + {x^3}...\infty } \right)$ is
$S\left( {1 - x} \right) = 1 + \dfrac{{2x}}{{1 - x}} \\ \Rightarrow S\left( {1 - x} \right) = \dfrac{{1 + x}}{{1 - x}} \\ \Rightarrow S = \dfrac{{1 + x}}{{{{\left( {1 - x} \right)}^2}}} \\$
Therefore, option C is correct.

Note: We can only apply the formula of infinite terms of series only when the $\left| r \right| < 1$, where $r$ is the common ratio. When there are infinite terms, the sum of G.P is ${S_\infty } = \dfrac{a}{{1 - r}}$, where $r$ is the common ratio, $a$ is the first term of the G.P.