If 0 1 1+x + 2 x 1+x^ 2 + 4 x^ 3 1+x^ 4 + 8 x^ 7 1+x^ 8 + =
Mathematics · JEE Advanced · NTA Exams — Limit, Continuity and Differentiability
If 0 < x < 1, then
\(\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{4 x^{3}}{1+x^{4}}+\frac{8 x^{7}}{1+x^{8}}+\ldots \infty=\)
\(\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{4 x^{3}}{1+x^{4}}+\frac{8 x^{7}}{1+x^{8}}+\ldots \infty=\)
- \(\frac{1}{1-x}\)
- \(\frac{\mathrm{x}}{1-\mathrm{x}}\)
- \(\frac{x}{1+x}\)
- \(\frac{1-x}{1+x}\)
Answer
(A) 1 1-x
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