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(C) Let $f(x) = \frac{2x^3+8x^2-2x-2}{(1-x)(1+x)(1-2x)}$.Using partial fractions,we write:$f(x) = 1 + \frac{A}{1-2x} + \frac{B}{1-x} + \frac{C}{1+x}$.

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$6(1+2^{10})$

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(C) Let $f(x) = \frac{2x^3+8x^2-2x-2}{(1-x)(1+x)(1-2x)}$.Using partial fractions,we write:$f(x) = 1 + \frac{A}{1-2x} + \frac{B}{1-x} + \frac{C}{1+x}$.Equating coefficients after simplification:$2x^3+8x^2-2x-2 = (1-x^2)(1-2x) + A(1-x^2) + B(1+x)(1-2x) + C(1-x)(1-2x)$.Solving for constants,we get $A = -2, B = -3, C = 1$.Thus,$f(x) = 1 - 2(1-2x)^{-1} - 3(1-x)^{-1} + (1+x)^{-1}$.Expanding using the binomial series $(1-z)^{-1} = \sum_{n=0}^{\infty} z^n$:$f(x) = 1 - 2 \sum_{n=0}^{\infty} (2x)^n - 3 \sum_{n=0}^{\infty} x^n + \sum_{n=0}^{\infty} (-x)^n$.The coefficient of $x^{10}$ $(l)$ is $-2(2^{10}) - 3(1) + 1 = -2^{11} - 2$.The constant term $(m)$ is $1 - 2(1) - 3(1) + 1 = -3$.Wait,re-evaluating the partial fraction decomposition:$f(x) = \frac{2x^3+8x^2-2x-2}{-(2x-1)(x^2-1)} = \frac{2x^3+8x^2-2x-2}{-(2x^3-x^2-2x+1)} = -1 + \frac{7x^2-1}{-(2x^3-x^2-2x+1)}$.Correct decomposition: $f(x) = 1 - \frac{2}{1-2x} - \frac{3}{1-x} + \frac{1}{1+x}$.$l = -2(2^{10}) - 3 + 1 = -2^{11} - 2$.$m = 1 - 2 - 3 + 1 = -3$.Given the options,$lm = (-2^{11}-2)(-3) = 3(2^{11}+2) = 6(2^{10}+1)$.

For $|x| < \frac{1}{2}$,if the coefficient of $x^{10}$ and the constant term in the expansion of $\frac{2 x^3+8 x^2-2 x-2}{(1-x)(1+x)(1-2 x)}$ in powers of $x$ are $l$ and $m$ respectively,then $lm=$

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