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