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Question
If the coefficient of second, third and fourth terms in the expansion of (1 + x)2n are in A.P. Show that 2n2 – 9n + 7 = 0.
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Solution
Given expression = (1 + x )2n
Coefficient of second term = 2nC1
Coefficient of third term = 2nC2
And coefficient of fourth term = 2nC3
As the given condition
2nC1. 2nC2 and 2nC3 are in A.P.
∴ 2nC2 – 2nC1 = 2nC3 – 2nC2
⇒ `2 * ""^(2n)"C"_2 = ""^(2n)"C"_1 + ""^(2n)"C"_3`
⇒ `2 * (2n!)/(2!(2n - 2)!) = (2n!)/((2n - 1)!) + (2n!)/(3!(2n - 3)!)`
⇒ `2[(2n(2n - 1)(2n - 2)!)/(2 xx 1 xx (2n - 2)!)] = (2n(2n - 1)!)/((2n - 1)!) + (2n(2n - 1)(2n - 2)(2n - 3)!)/(3 xx 2 xx 1 xx (2n - 3)!)`
⇒ n(2n – 1) = `n + (n(2n - 1)(2n - 2))/6`
⇒ 2n – 1 = `1 + ((2n - 1)(2n - 2))/6`
⇒ 12n – 6 = 6 + 4n2 – 4n – 2n + 2
⇒ 12n – 12 = 4n2 – 6n + 2
⇒ 4n2 – 6n – 12n + 2 + 12 = 0
⇒ 4n2 – 18n + 14 = 0
⇒ 2n2 – 9n + 7 = 0
Hence proved.
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