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Question

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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.${S_\infty } = \dfrac{{1 + x}}{{{{\left( {1 - x} \right)}^3}}}$

Answer
Verified

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}}} \\

$