The coefficient of $x^{10}$ in the expansion of $(1 + x)^{2} (1 + x^{2})^{3} (1 + x^{3})^{4}$ is qual to:

52

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56

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50

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44

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Solution

The correct option is A $52$
We know,
$(a + b)^{2} = a^{2} + b^{2} + 2 a b$
$(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$
$(a + b)^{4} = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4}$

$(1 + x)^{2} (1 + x^{2})^{3} (1 + x^{3})^{4}$
$\Rightarrow$ $(1 + x^{2} + 2 x) (1 + 3 x^{2} + 3 x^{4} + x^{6}) (1 + 4 x^{3} + 6 x^{6} + 4 x^{9} + x^{12})$
$\Rightarrow$ $(1 + 3 x^{2} + 3 x^{4} + x^{6} + x^{2} + 3 x^{4} + 3 x^{6} + x^{8} + 2 x + 6 x^{3} + 6 x^{5} + 2 x^{7}) (1 + 4 x^{3} + 6 x^{6} + 4 x^{9} + x^{12})$
$\Rightarrow$ $(1 + 2 x + 4 x^{2} + 6 x^{4} + 6 x^{5} + 4 x^{6} + 2 x^{7} + x^{8}) (1 + 4 x^{3} + 6 x^{6} + 4 x^{9} + x^{12})$
Now, coefficient of $x^{10}$ $= (4 \times 2) + (6 \times 6) + (2 \times 4)$
$= 8 + 36 + 8$
$= 52$

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