If the coefficient of second, third and fourth terms in the expansion of (1 + x)^{2n} are in A.P. Show that 2n^{2} – 9n + 7 = 0.

#### Solution

Given expression = (1 + x )^{2n}

Coefficient of second term = ^{2n}C_{1}

Coefficient of third term = ^{2n}C_{2}

And coefficient of fourth term = ^{2n}C_{3}

As the given condition

^{2n}C_{1}. ^{2n}C_{2} and ^{2n}C_{3} are in A.P.

∴ ^{2n}C_{2} – ^{2n}C_{1} = ^{2n}C_{3} – ^{2n}C_{2}

⇒ `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 + 4n^{2} – 4n – 2n + 2

⇒ 12n – 12 = 4n^{2} – 6n + 2

⇒ 4n^{2} – 6n – 12n + 2 + 12 = 0

⇒ 4n^{2} – 18n + 14 = 0

⇒ 2n^{2} – 9n + 7 = 0

Hence proved.