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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...

If the coefficients of second, third and fourth term in the expansion of (1+x)2n(1 + x)^{2n} are in A.P., then 2n29n+72n^{2} - 9n + 7 is equal to

A

–1

B

0

C

1

D

3/2

Answer

0

Explanation

Solution

T2=2nC1T_{2} =^{2n} ⥂ C_{1}, T3=2nC2,T4=2nC3T_{3} =^{2n} ⥂ C_{2},T_{4} =^{2n} ⥂ C_{3} are in A.P. then,

2.2nC2=2nC1+2nC32.^{2n} ⥂ C_{2} =^{2n} ⥂ C_{1} +^{2n} ⥂ C_{3}

2.2n(.2n1)2.1=2n1+2n(2n1)(2n2)3.2.12.\frac{2n(.2n - 1)}{2.1} = \frac{2n}{1} + \frac{2n(2n - 1)(2n - 2)}{3.2.1}

On solving, 2n29n+7=02n^{2} - 9n + 7 = 0