the coefficients of 2nd, third and fourth terms in the expansion of (1+n)^2n are in AP.Prove that 2n^2-9n+7=0 Share with your friends Share 12 Vipin Verma answered this Expansion of (1+x)2n=2nC0x2n+2nC1x2n-1+2nC2x2n-2+.....Let Tn denotes the nth term then we haveT2=2nC1x2n-1T3=2nC2x2n-2T4=2nC3x2n-3Since coefficients of 2nd,3rd and 4th term are in APSo,2×2nC2=2nC1+2nC3⇒2(2n)!(2n-2)!2!=(2n)!(2n-1)!1!+(2n)!(2n-3)!3!⇒1(2n-2)=1(2n-1)(2n-2)+16⇒1(2n-2)=6+(2n-1)(2n-2)6(2n-1)(2n-2)⇒6(2n-1)=6+4n2-6n+2⇒12n-6=4n2-6n+8⇒4n2-6n+8-12n+6=0⇒4n2-18n+14=0Dividing by 2 we get⇒2n2-9n++7=0 Proved 34 View Full Answer Manjaree Agarwal answered this T2=2nC1 x2n-1y T3=2nC2x2n-2y2 T4=2nC3x2n-3y32nC1 +2nC3 =2*2nC2 (by the A.P. formula)(2n)! + (2n)! = 2* (2n)!(2n-1)! 3!(2n-3)! (2n-2)!2!(2n)! { 1 + 1 } = 2 (by taking common)(2n-3)! {(2n-2)(2n-3) 3*2 } (2n-2)(2n-3)! * 26+4n2-6n+2 = 1 (2n-2)(2n-1)6 (2n-2)4n2-6n+8=12n-62n2-3n+4=6n-3 (dividing the equation by 2)2n2-3n+4-6n+3=02n2-9n+7=0 23