If the coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)2n are in AP, show that 2n2 – 9n + 7 = 0.

For (1 + x)2n


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


We have,


For the 2nd term, r=1




………


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


For the 3rd term, r=2





……….(n! = n. (n-1)!)



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


For the 4th term, r=3





……….(n! = n. (n-1)!)





Therefore, the coefficient of 3rd term


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


Therefore,


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



Dividing throughout by (2n),




• 3 (2n-1) = 3 + (2n-1)(n-1)


• 6n – 3 = 3 + (2n2 - 2n – n + 1)


• 6n – 3 = 3 + 2n2 - 3n + 1


• 3 + 2n2 - 3n + 1 - 6n + 3 = 0


• 2n2 - 9n + 7 = 0


Conclusion : If the coefficients of 2nd, 3rd and 4th terms of (1 + x)2n are in A.P. then 2n2 - 9n + 7 = 0


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