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Q.

The coefficient of x5  in the expansion of (x2x2)5  is

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a

83

b

82

c

81

d

0

answer is C.

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Detailed Solution

(x2x2)5=(x22x+x2)5

=(x(x2)+1(x2))5

=(x2)5(x+1)5

=(1+x)5(x2)5

=[5C01+5C1x+5C2x2+5C3x3+5C4x4+5C5x5]

×[5C0.x55C1x4.2+5C2x3.225C3x223+5C4x.245C525]

=[5C0×5C0+5C1×(5C1(2))+5C2×5C2×22+5C3×(5C3)23+5C4×5C4.24+5C5×5C52]

=150+400800+40032

=81

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