Let a,b,c such that 1(1−x)(1−2x)(1−3x)=a1−x+b1−2x+c1−3x, then the value of a1+b3+c5=

1=a(1−2x)(1−3x)+b(1−x)(1−3x)+c(1−x)(1−2x)Put x=1⇒a=12 x=12⇒b=−4 x=13⇒c=92 ∴a1+b3+c5=115

Let a,b,c such that 1(1−x)(1−2x)(1−3x)=a1−x+b1−2x+c1−3x, then the value of a1+b3+c5=

1=a(1−2x)(1−3x)+b(1−x)(1−3x)+c(1−x)(1−2x)Put x=1⇒a=12 x=12⇒b=−4 x=13⇒c=92 ∴a1+b3+c5=115

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

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$\frac{1}{6}$

$\frac{1}{5}$

$\frac{1}{15}$

$\frac{1}{3}$

answer is A.

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Detailed Solution

$1 = a (1 - 2 x) (1 - 3 x) + b (1 - x) (1 - 3 x) + c (1 - x) (1 - 2 x)$
Put $x = 1 \Rightarrow a = \frac{1}{2}$
$x = \frac{1}{2} \Rightarrow b = - 4 x = \frac{1}{3} \Rightarrow c = \frac{9}{2} ∴ \frac{a}{1} + \frac{b}{3} + \frac{c}{5} = \frac{1}{15}$