Answer :

For (1 + x)^{2n}

a=1, b=x and N=2n

We have,

For the 2^{nd} term, r=1

………

Therefore, the coefficient of 2^{nd} term = (2n)

For the 3^{rd} term, r=2

……….(n! = n. (n-1)!)

Therefore, the coefficient of 3^{rd} term = (n)(2n-1)

For the 4^{th} term, r=3

……….(n! = n. (n-1)!)

Therefore, the coefficient of 3^{rd} term

As the coefficients of 2^{nd}, 3^{rd} and 4^{th} terms are in A.P.

Therefore,

2×coefficient of 3^{rd} term = coefficient of 2^{nd} term + coefficient of the 4^{th} term

Dividing throughout by (2n),

• 3 (2n-1) = 3 + (2n-1)(n-1)

• 6n – 3 = 3 + (2n^{2} - 2n – n + 1)

• 6n – 3 = 3 + 2n^{2} - 3n + 1

• 3 + 2n^{2} - 3n + 1 - 6n + 3 = 0

• 2n^{2} - 9n + 7 = 0

__Conclusion__ : If the coefficients of 2^{nd}, 3^{rd} and 4^{th} terms of (1 + x)^{2n} are in A.P. then 2n^{2} - 9n + 7 = 0

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