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Let $a, b$, and $c$ be 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 $\frac{a}{1}+\frac{b}{3}+\frac{c}{5}$ is equal to
Options:
$\frac{1}{(1-x)(1-2 x)(1-3 x)}=\frac{a}{1-x}+\frac{b}{1-2 x}+\frac{c}{1-3 x}$
then $\frac{a}{1}+\frac{b}{3}+\frac{c}{5}$ is equal to
Solution:
2044 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{15}$
We have,
$1=a(1-2 x)(1-3 x)+b(1-x)(1-3 x)$$+c(1-x)(1-2 x)$
Put $x=\frac{1}{2}$
$1=0+b\left(1-\frac{1}{2}\right)\left(1-\frac{3}{2}\right)+0$
$\Rightarrow \quad 1=b\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right) \Rightarrow b=-4$
Put $x=1$,
$1=a(1-2)(1-3)+0+0$
$\Rightarrow \quad 1=a(-1)(-2) \Rightarrow a=\frac{1}{2}$
Put $x=\frac{1}{3}$,
$1=0+0+c\left(1-\frac{1}{3}\right)\left(1-\frac{2}{3}\right)$
$\Rightarrow \quad 1=c\left(\frac{2}{3}\right)\left(\frac{1}{3}\right) \Rightarrow c=\frac{9}{2}$
Now, $\frac{a}{1}+\frac{b}{3}+\frac{c}{5}=\frac{1}{2}+\frac{(-4)}{3}+\frac{9}{5 \cdot 2}$
$=\frac{1}{2}-\frac{4}{3}+\frac{9}{10}$
$=\frac{15-40+27}{30}=\frac{2}{30}=\frac{1}{15}$
$1=a(1-2 x)(1-3 x)+b(1-x)(1-3 x)$$+c(1-x)(1-2 x)$
Put $x=\frac{1}{2}$
$1=0+b\left(1-\frac{1}{2}\right)\left(1-\frac{3}{2}\right)+0$
$\Rightarrow \quad 1=b\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right) \Rightarrow b=-4$
Put $x=1$,
$1=a(1-2)(1-3)+0+0$
$\Rightarrow \quad 1=a(-1)(-2) \Rightarrow a=\frac{1}{2}$
Put $x=\frac{1}{3}$,
$1=0+0+c\left(1-\frac{1}{3}\right)\left(1-\frac{2}{3}\right)$
$\Rightarrow \quad 1=c\left(\frac{2}{3}\right)\left(\frac{1}{3}\right) \Rightarrow c=\frac{9}{2}$
Now, $\frac{a}{1}+\frac{b}{3}+\frac{c}{5}=\frac{1}{2}+\frac{(-4)}{3}+\frac{9}{5 \cdot 2}$
$=\frac{1}{2}-\frac{4}{3}+\frac{9}{10}$
$=\frac{15-40+27}{30}=\frac{2}{30}=\frac{1}{15}$
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