Let a, b, c such that 11-x1-2x1-2x=a1-x+b1-2x+c1-3x, a1+b3+c5=

1(1-x)(1-2x)(1-3x)=a1-x+b1-2x+c1-3xx=2 ⇒ 1(-1)(-3)(-5)=a-1+b-3+c-5 ⇒a1+b3+c5=115

Let a, b, c such that 11-x1-2x1-2x=a1-x+b1-2x+c1-3x, a1+b3+c5=

1(1-x)(1-2x)(1-3x)=a1-x+b1-2x+c1-3xx=2 ⇒ 1(-1)(-3)(-5)=a-1+b-3+c-5 ⇒a1+b3+c5=115

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$Let a, b, c such that \frac{1}{(1 - x) (1 - 2 x) (1 - 2 x)} = \frac{a}{1 - x} + \frac{b}{1 - 2 x} + \frac{c}{1 - 3 x}, \frac{a}{1} + \frac{b}{3} + \frac{c}{5} =$

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$\frac{1}{6}$

$\frac{1}{15}$

$\frac{1}{5}$

$\frac{1}{3}$

answer is A.

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Detailed Solution

$\begin{array}{rcl} \frac{1}{(1 - x) (1 - 2 x) (1 - 3 x)} & = & \frac{a}{1 - x} + \frac{b}{1 - 2 x} + \frac{c}{1 - 3 x} \\ x & = & 2 \Rightarrow \frac{1}{(- 1) (- 3) (- 5)} = \frac{a}{- 1} + \frac{b}{- 3} + \frac{c}{- 5} \\ \Rightarrow & \frac{a}{1} + \frac{b}{3} + \frac{c}{5} = \frac{1}{15} \end{array}$