If the number of terms in the expansion of (x+y+z)n is 231, then the value of n is
A
20
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B
30
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C
31
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D
21
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Solution
The correct option is A20 Given expansion (x+y+z)n
The number of terms 231=n+3−1C3−1⇒231=n+2C2⇒(n+2)(n+1)2=231⇒n2+3n+2=462⇒n2+3n−460=0⇒(n+23)(n−20)=0⇒n=20(∵n∈N)