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

 

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+2nC32(2n)!(2n-2)!2!=(2n)!(2n-1)!1!+(2n)!(2n-3)!3!1(2n-2)=1(2n-1)(2n-2)+161(2n-2)=6+(2n-1)(2n-2)6(2n-1)(2n-2)6(2n-1)=6+4n2-6n+212n-6=4n2-6n+84n2-6n+8-12n+6=04n2-18n+14=0Dividing by 2 we get2n2-9n++7=0       Proved

 

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T2=2nC1 x2n-1y T3=2nC2x2n-2y2 T4=2nC3x2n-3y3

2nC1 +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)! * 2

6+4n2-6n+2 = 1

(2n-2)(2n-1)6 (2n-2)

4n2-6n+8=12n-6

2n2-3n+4=6n-3 (dividing the equation by 2)

2n2-3n+4-6n+3=0

2n2-9n+7=0

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