The coefficient of $x^{5}$ in the expansion of $(1 + x^{2})^{5} (1 + x)^{4}$ is ?

61

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z e r o

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Solution

The correct option is C $z e r o$
according to binomial expansion,

$(1 + x^{2})^{5} = 5 C_{0} \times 1 + 5 C_{1} (x^{2}) + 5 C_{2} (x^{2})^{2} + 5 C_{3} (x^{2})^{3} + 5 C_{4} (x^{2})^{4} + 5 C_{5} (x^{2})^{5}$

$= 1 + 5 x^{2} + 10 x^{4} + 10 x^{6} + 5 x^{8} + X^{0} ⟹ (1)$

$= (1 + x)^{4} = 1 + 4 C_{1} x + 4 C_{2} x^{2} + 4 C_{3} x^{3} + 4 C_{4} x^{4} ⟹ (2)$

from 1 and 2 we can see that coefficient of 5 is not there

so $x^{5}$ is zero

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Similar questions

x^{5}

{(1 + x^{2})}^{5} (1 + x)^{4}

60

x^{5}

(1 + x^{2})^{5} (1 + x)^{4}

The coefficient of $x^{5}$ in the expansion of

$(1 + x^{2})^{5}$ $(1 + x)^{4}$ is

x^{5}

(1 + x^{2})^{5} (1 + x)^{4}

x^{5}

(1 + x^{2})^{5} (1 + x)^{4}

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