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Question

Number of terms free from radical sign in the expansion of (1+31/3+71/2)10 is

A
4
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B
5
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C
6
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D
8
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Solution

The correct option is D 8
(1+(313+712))10
Let
(313+712)=x
Then the above expression reduces to,
(1+x)10
Now writing the general term, we get
Tr+1=10Crxr
=10Cr(313+712)r.
Hence we get one rational term for each
r=0,2,3,4,6,8,9,10
Hence in total 8 terms.

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