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(A) The binomial expansion for $(1+y)^n$ where $|y| < 1$ is $1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \ldots$Since $0 < x < 1$,we h

Vedclass · Accepted Answer

$1+\frac{1}{2 x}-\frac{1}{2 !}\left(\frac{1}{2 x}\right)^2+\frac{1 \cdot 3}{3 !}\left(\frac{1}{2 x}\right)^3-\frac{1 \cdot 3 \cdot 5}{4 !}\left(\frac{1}{2 x}\right)^4+\ldots \infty$

Vedclass · Answer

(A) The binomial expansion for $(1+y)^n$ where $|y| Since $0  1$. Thus,we rewrite the expression as $\left(\frac{1}{x}\right)^{\frac{1}{2}} (1+x)^{\frac{1}{2}} = \frac{1}{\sqrt{x}} (1+x)^{\frac{1}{2}}$.Expanding $(1+x)^{\frac{1}{2}} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \ldots$Multiplying by $\frac{1}{\sqrt{x}}$,we get $\frac{1}{\sqrt{x}} + \frac{\sqrt{x}}{2} - \frac{x\sqrt{x}}{8} + \ldots$However,if we expand $\left(1+\frac{1}{x}\right)^{\frac{1}{2}}$ directly as $\left(\frac{1}{x}\right)^{\frac{1}{2}} (x+1)^{\frac{1}{2}}$,it is not valid for $|\frac{1}{x}| > 1$. The question implies the standard expansion form for $\left(1+\frac{1}{x}\right)^{\frac{1}{2}}$ which is $1 + \frac{1}{2x} - \frac{1}{8x^2} + \ldots$ which matches option $A$.

For $0 < x < 1$,the expansion of $\left(1+\frac{1}{x}\right)^{\frac{1}{2}}$ is

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