If ${(1 + 2 x + 3 x^{2})}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{20} x^{20}$ , then

a_{1} = 20

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a_{2} = 210

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a_{4} = 8085

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a_{20} = 2^{2} \times 3^{7} \times 7

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Solution

The correct options are
A $a_{1} = 20$
C $a_{4} = 8085$
D $a_{2} = 210$
Coefficient of x will be coefficient of $(2 x)^{1} (1)^{9} (3 x^{2})^{0}$
$= \frac{10!}{1! .9!} .2$
$= 20$
Coefficient of $x^{2}$ will be coefficient of $(2 x)^{2} (1)^{8} (3 x^{2})^{0}$ + $(2 x)^{0} 1^{9} (3 x^{2})^{1}$
$= \frac{10!}{2! .8!} .4 + \frac{10!}{9! .1!} .3$
$= 180 + 30$
$= 210$
Coefficient of $x^{4}$ will be sum of the coefficients of $(2 x)^{4} (3 x^{2})^{0} 1^{6}$ + $(2 x)^{2} (3 x^{2})^{1} 1^{7}$ + $(2 x)^{0} (3 x^{2})^{2} 1^{8}$
$= 16 \frac{10!}{4! .6!} + 12 \frac{10!}{7! .2!} + 9 \frac{10!}{2! .8!}$
$= 420 + 4320 + 405$
$= 8085$
The coefficient of $x^{20}$ will be the coefficient of $(3 x^{2})^{10}$
$= 3^{10}$

Suggest Corrections

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