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Question

If $$x<1$$,  then $$\displaystyle\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\cdots\infty=$$


A
11+x
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B
11x
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C
11x
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D
11+x
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Solution

The correct option is B $$\displaystyle\frac{1}{1-x}$$
The given series is in the form
$$\displaystyle\frac{{f_1}^\prime(x)}{{f_1}(x)}+\frac{{f_2}^\prime(x)}{{f_2}(x)}+\frac{{f_3}^\prime(x)}{{f_3}(x)}+\cdots\infty$$
Then consider the product $${f_1}(x).{f_2}(x).{f_3}(x)\cdots{f_n}(x)$$. Now,
$$(1-x)(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^{ n-1}})$$ ---------(1)
$$=(1-x^2)(1+x^2)(1+x^4)\cdots(1+x^{2^{ n-1}})$$
$$=(1-x^4)(1+x^4)\cdots(1+x^{2^{ n-1}})$$
$$\:\vdots$$
$$=(1-x^{2^{n-1}})(1+x^{2^{ n-1}})$$
$$=1-x^{2^{ n}}$$
Now, when $$n\rightarrow\infty$$, $$x^{2^{ n-1}}\rightarrow 0\:(\because x<1)$$.
Therefore, taking $$n\rightarrow\infty$$, in (1), we get
$$(1-x)(1+x)(1+x^2)(1+x^4)\cdots=1$$
Taking logarithm, we get
$$\log{(1-x)}+\log{(1+x)}+\log{(1+x^2)}+\log{(1+x^4)}+\cdots=0$$
Differentiating w.r.t. $$x$$, we get
$$\displaystyle -\frac{1}{1-x}+\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\cdots=0$$
or $$\displaystyle\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\cdots\infty=\frac{1}{1-x}$$

Mathematics

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