When |x|

Question

(B) Given the expression $f(x) = \left(\frac{2-x}{1+2x}\right)^2 = (2-x)^2 (1+2x)^{-2}$.Expanding $(2-x)^2 = 4 - 4x + x^2$.Using the binomial expansio

Vedclass · Accepted Answer

$2640$

Vedclass · Answer

(B) Given the expression $f(x) = \left(\frac{2-x}{1+2x}ight)^2 = (2-x)^2 (1+2x)^{-2}$.Expanding $(2-x)^2 = 4 - 4x + x^2$.Using the binomial expansion for $(1+2x)^{-2} = \sum_{n=0}^{\infty} \binom{-2}{n} (2x)^n = \sum_{n=0}^{\infty} (-1)^n (n+1) (2x)^n = \sum_{n=0}^{\infty} (-1)^n (n+1) 2^n x^n$.Thus,$f(x) = (4 - 4x + x^2) \sum_{n=0}^{\infty} (-1)^n (n+1) 2^n x^n$.The coefficient of $x^6$ is obtained by:$4 	imes [	ext{coeff of } x^6] - 4 	imes [	ext{coeff of } x^5] + 1 	imes [	ext{coeff of } x^4]$.$= 4 	imes [(-1)^6 (6+1) 2^6] - 4 	imes [(-1)^5 (5+1) 2^5] + 1 	imes [(-1)^4 (4+1) 2^4]$.$= 4 	imes (7 	imes 64) - 4 	imes (-6 	imes 32) + (5 	imes 16)$.$= 4 	imes 448 + 4 	imes 192 + 80$.$= 1792 + 768 + 80 = 2640$.

When $|x| < \frac{1}{2}$,the coefficient of $x^6$ in the expansion of $\left(\frac{2-x}{1+2x}\right)^2$ is

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