Question

Find \dfrac{1}{{1 + {x^2}}}\left\\{ {1 + \dfrac{{2x}}{{1 + {x^2}}} + {{\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)}^2} + \ldots \ldots \ldots + \infty } \right\\}

Answer 1

In the given question, we are asked to find the value of the given expression. There is a series inside a bracket. If we observe it carefully, we will find that the given series is a geometric progression i.e. G.P. We will solve the G.P. first using the applicable identity and then multiply it by the term outside the bracket to get the desired result.
Formula used:
${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
Sum of infinite terms of a G.P. is given by ${S_\infty } = \dfrac{a}{{1 - r}}$,
where, $a$ is the first term of a G.P. and $r$ is the common ratio, $0 < r < 1$.

Complete step by step answer:
To find: \dfrac{1}{{1 + {x^2}}}\left\\{ {1 + \dfrac{{2x}}{{1 + {x^2}}} + {{\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)}^2} + \ldots \ldots \ldots + \infty } \right\\}
Observing the series inside the bracket, we find it to be a G.P. with $a = 1$ and $r = \dfrac{{2x}}{{1 + {x^2}}}$. Now, we will solve the G.P. first using the formula of sum of infinite terms of a G.P.
Let $A = \,1 + \dfrac{{2x}}{{1 + {x^2}}} + {\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)^2} + \ldots \ldots \ldots + \infty$
Sum of infinite terms of a G.P. is given by ${S_\infty } = \dfrac{a}{{1 - r}}$,
where, $a$ is the first term of a G.P. and $r$ is the common ratio, $0 < r < 1$.
$\therefore A = \dfrac{1}{{1 - \dfrac{{2x}}{{1 + {x^2}}}}}$
Simplifying it, we get,

$$A = \dfrac{1}{{\dfrac{{1 + {x^2} - 2x}}{{1 + {x^2}}}}} \\\ \Rightarrow A = \dfrac{{1 + {x^2}}}{{1 + {x^2} - 2x}} \\\$$

Using the identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$,
$A = \dfrac{{1 + {x^2}}}{{{{\left( {1 - x} \right)}^2}}}$
As we have solved the G.P, now we will multiply by the term given outside the bracket.
Therefore, the given expression will become,
\dfrac{1}{{1 + {x^2}}}\left\\{ {\dfrac{{1 + {x^2}}}{{{{\left( {1 - x} \right)}^2}}}} \right\\}
Cancelling the common term in numerator and denominator,
$= \dfrac{1}{{{{\left( {1 - x} \right)}^2}}}$
Hence, \dfrac{1}{{1 + {x^2}}}\left\\{ {1 + \dfrac{{2x}}{{1 + {x^2}}} + {{\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)}^2} + \ldots \ldots \ldots + \infty } \right\\} = \dfrac{1}{{{{\left( {1 - x} \right)}^2}}}.

Note: While solving such types of questions, it is important that we must understand the different types of series, usually, arithmetic progression and geometric progression and the related formulae. Do not forget to multiply both the terms, after solving for G.P, in the end as it will lead to incomplete solution and hence, wrong answer.