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Question

The coefficient of x1012 in the expansion of (1+xn+x253)10, (where n22 is any positive integer), is :

A
1
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B
10C4
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C
4n
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D
253C4
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Solution

The correct option is B 10C4
Tr+1=10Cr(1+xn)10r(x253)r
253r=1012
r=4
Coefficient is 10C4 .

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