For 0 < x < 1, the expansion of (1+1/x)^{1/2} is

Correct Answer is : 1√x+12√x−12ix√x22+1.33!x2√x23−.........∞

1+1/{2x}-1/{2!}(1/{2x})^2+1.3/{3!}(1/{2x})^3-\frac{1.3.5}{4!}(1/{2x})^4+.....\infty

1/\sqrt{x}+1/2\sqrt{x}-1/{2i}{x\sqrt{x}/2^2+1.3/{3!}{{x^2\sqrt{x}/2^3-.........\infty

1+1/\sqrt{x}+1/2x\sqrt{x}+1/{2i}{x^2\sqrt{x}/2^3+\frac{1.3}{3!}{x^3\sqrt{x}/2^4+........\infty

1/\sqrt{x}+1/{2x\sqrt{x}-1/{2i}(1/{2x})^2 1/\sqrt{x}+\frac{1.3}{3!}(1/{2x})^3 1/\sqrt{x}-.......\infty

TS EAMCET 10-Sep-2020 Shift 1 Solved Paper

Section: Mathematics

Question : 18 of 160

Marks: +1, -0

For 0<x<1, the expansion of (1+

)

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