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Question

If the number of terms in (a+b)n2+3 and (c+d)3n+4 are same, where a,b,c,d0,nN, then the number of possible value(s) of n is

A
0
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B
1
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C
2
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D
More than 2 but finite.
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Solution

The correct option is A 0
Given expansions are (a+b)n2+3 and (c+d)3n+4
As the number of terms are same, so
n2+3+1=3n+4+1n23n1=0n=3±9+42
As nN, so there is no possible value of n.

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