TS EAMCET · Maths · Binomial Theorem
For \(0 < x < 1\), the expansion of \(\left(1+\frac{1}{x}\right)^{\frac{1}{2}}\) is
- A \(1+\frac{1}{2 x}-\frac{1}{2 !}\left(\frac{1}{2 x}\right)^2+\frac{1 \cdot 3}{3 !}\left(\frac{1}{2 x}\right)^3-\frac{1 \cdot 3 \cdot 5}{4 !}\left(\frac{1}{2 x}\right)^4+\ldots \infty\)
- B \(\frac{1}{\sqrt{x}}+\frac{1}{2} \sqrt{x}-\frac{1}{2 !} \frac{x \sqrt{x}}{2^2}+\frac{1 \cdot 3}{3 !} \frac{x^2 \sqrt{x}}{2^3}-\ldots . \infty\)
- C \(1+\frac{1}{\sqrt{x}}+\frac{1}{2} x \sqrt{x}+\frac{1}{2 !} \frac{x^2 \sqrt{x}}{2^3}+\frac{1 \cdot 3}{3 !} \frac{x^3 \sqrt{x}}{2^4}+\ldots \infty\)
- D \(\frac{1}{\sqrt{x}}+\frac{1}{2 x \sqrt{x}}-\frac{1}{2 !}\left(\frac{1}{2 x}\right)^2 \frac{1}{\sqrt{x}}+\frac{1 \cdot 3}{3 !}\left(\frac{1}{2 x}\right)^3 \frac{1}{\sqrt{x}}-\ldots \ldots \infty\)
Answer & Solution
Correct Answer
(A) \(1+\frac{1}{2 x}-\frac{1}{2 !}\left(\frac{1}{2 x}\right)^2+\frac{1 \cdot 3}{3 !}\left(\frac{1}{2 x}\right)^3-\frac{1 \cdot 3 \cdot 5}{4 !}\left(\frac{1}{2 x}\right)^4+\ldots \infty\)
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