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Question 84 of 95

Q.Expand (x+2)−2/3(x+2)^{-2/3} in powers of xx.

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2022Subjective· 3mImportance★★★★★
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Writing (x+2)−2/3=2−2/3(1+x/2)−2/3(x+2)^{-2/3}=2^{-2/3}(1+x/2)^{-2/3} and applying the general binomial expansion gives 2−2/3[1−x3+5x236−5x381+⋯ ]2^{-2/3}\left[1-\frac{x}{3}+\frac{5x^2}{36}-\frac{5x^3}{81}+\cdots\right].

(x+2)−2/3=[2(1+x2)]−2/3=2−2/3(1+x2)−2/3(x+2)^{-2/3} = \left[2\left(1+\dfrac{x}{2}\right)\right]^{-2/3} = 2^{-2/3}\left(1+\dfrac{x}{2}\right)^{-2/3}, valid for ∣x2∣<1\left|\dfrac{x}{2}\right|<1.

Using the general binomial series (1+t)n=1+nt+n(n−1)2!t2+n(n−1)(n−2)3!t3+⋯(1+t)^n = 1+nt+\dfrac{n(n-1)}{2!}t^2+\dfrac{n(n-1)(n-2)}{3!}t^3+\cdots with n=−23n=-\dfrac23, t=x2t=\dfrac{x}{2}:

Coefficient of tt: n=−23n=-\dfrac23.

Coefficient of t2t^2: n(n−1)2=(−2/3)(−5/3)2=10/92=59\dfrac{n(n-1)}{2} = \dfrac{(-2/3)(-5/3)}{2} = \dfrac{10/9}{2}=\dfrac59.

Coefficient of t3t^3: n(n−1)(n−2)6=(−2/3)(−5/3)(−8/3)6=−80/276=−4081\dfrac{n(n-1)(n-2)}{6} = \dfrac{(-2/3)(-5/3)(-8/3)}{6} = \dfrac{-80/27}{6}=-\dfrac{40}{81}.

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