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Question

Coefficient of the term independent of x in the expansion of
x+1x23x13+1x1xx1210 is

A
105
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B
210
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C
310
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D
180
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Solution

The correct option is D 210
Let x=t6
Hence,
(t6+1t4t2+1t61t6t3)10
=((t2+1)(t4t2+1)t4t2+1(t31)(t3+1)t3(t31))10
=(t2+111t3)10
=(t21t3)10
Tr+1=(1)r10Crt202rt3r
Hence term independent of t will be given by
205r=0
r=4
Hence the coefficient is 10C4 =210.

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