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Question: The coefficient of \(x^{n}\) in the expansion of \(\left( \frac{1}{1 - x} \right)\left( \frac{1}{3 -...

The coefficient of xnx^{n} in the expansion of (11−x)(13−x)\left( \frac{1}{1 - x} \right)\left( \frac{1}{3 - x} \right) is

A

3n+1−12.3n+1\frac{3^{n + 1} - 1}{2.3^{n + 1}}

B

3n+1−13n+1\frac{3^{n + 1} - 1}{3^{n + 1}}

C

2(3n+1−13n+1)2\left( \frac{3^{n + 1} - 1}{3^{n + 1}} \right)

D

None

Answer

3n+1−12.3n+1\frac{3^{n + 1} - 1}{2.3^{n + 1}}

Explanation

Solution

1(1−x)(3−x)=(1−x)−1(3−x)−1\frac{1}{(1 - x)(3 - x)} = (1 - x)^{- 1}(3 - x)^{- 1}= 3−1(1−x)−1(1−x3)−13^{- 1}(1 - x)^{- 1}\left( 1 - \frac{x}{3} \right)^{- 1}

=13[1+x+x2+.....xn][1+x3+x232+.....+xn−13n−1+xn3n]\frac{1}{3}\lbrack 1 + x + x^{2} + .....x^{n}\rbrack\left\lbrack 1 + \frac{x}{3} + \frac{x^{2}}{3^{2}} + ..... + \frac{x^{n - 1}}{3^{n - 1}} + \frac{x^{n}}{3^{n}} \right\rbrackCoefficient of xn=13n+1+13n+13n−1+.....(n+1) termsx^{n} = \frac{1}{3^{n + 1}} + \frac{1}{3^{n}} + \frac{1}{3^{n - 1}} + .....(n + 1)\text{ terms}

= 13n+1[3n+1−1]3−1=3n+1−12.3n+1\frac{1}{3^{n + 1}}\frac{\lbrack 3^{n + 1} - 1\rbrack}{3 - 1} = \frac{3^{n + 1} - 1}{2.3^{n + 1}}.

Trick: Put n=1,2,3......n = 1,2,3...... and find the coefficients of x,x2,x3......x,x^{2},x^{3}...... and comparing with the given option as :

Coefficient of x2x^{2} is = 133+132+131\frac{1}{3^{3}} + \frac{1}{3^{2}} + \frac{1}{3^{1}} = 133[33−1]3−1=1327\frac{1}{3^{3}}\frac{\lbrack 3^{3} - 1\rbrack}{3 - 1} = \frac{13}{27}; Which is given by option (1) 3n+1−12.(3n+1)=33−12.33=1327\frac{3^{n + 1} - 1}{2.(3^{n + 1})} = \frac{3^{3} - 1}{2.3^{3}} = \frac{13}{27}.