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Question

If the coefficients of second, third and fourth term in the expansion of (1+x)2n are in A.P, then 2n29n+7 is equal to

A
-1
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B
1
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C
3/2
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Solution

T2=2nC1x,T3=2nC2x2,T4=2nC3x3
Coefficient of T1,T2,T are in A.P.
2.2nC2=2nC1+2nC3
22n!2!(2n2)!=2n!(2n2)!+2n!3!(2n3)!
2.2n(2n1)2=2n+2n(2n1)(2n2)6
n(2n1)=n+(n)(2n1)(2n2)6
6(2n2n)=6n+4n36n2+2n
6n(2n1)=2n(2n23n+4)
6n3=2n23n+4
0=2n29n+72n29n+7=0

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