The coefficient of $x^{12} i n (x^{3} + x^{4} + x^{5} + x^{6} . . . . .)^{3}$ is
10

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Solution

The correct option is A 10
$(x^{3} + x^{4} + x^{5} + x^{6} . . . .^{3}$
$= x^{9} ((1 + x + x^{2} + . . .)^{3}$
$= x^{9} ((1 - x)^{- 1})^{3}$
$= x^{9} (1 - x)^{- 3}$
$= x^{9} [1 + (- 3) (- x) + \frac{(- 3) (- 3 - 1)}{2!} (- x)^{2} + \frac{(- 3) (- 3 - 1) (- 3 - 2)}{3!} (- x)^{3} + . . . . . .]$
$= x^{9} [1 + 3 x + 6 x^{2} + 10 x^{3} + . . . . . .]$
So, Coeffiecient of $x^{12} = 10$

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The coefficient of x12in(x3+x4+x5+x6.....)3 is 10

The correct option is A 10(x3+x4+x5+x6....3 =x9((1+x+x2+...)3 =x9((1−x)−1)3 =x9(1−x)−3 =x9[1+(−3)(−x)+(−3)(−3−1)2!(−x)2+(−3)(−3−1)(−3−2)3!(−x)3+......] =x9[1+3x+6x2+10x3+......] So, Coeffiecient of x12=10

The coefficient of $x^{12} i n (x^{3} + x^{4} + x^{5} + x^{6} . . . . .)^{3}$ is
10