Challenge!

(1 + x)^(2n)
= 1 + 2nx + [ 2n(2n - 1)/2 ]x^2 + [2n(2n-1)(2n-2) / 6] x^3 + ...

The coeff. of 2nd, 3rd, 4th terms in the expansion of (1+x)^(2n) are in A.P.
So 2n(2n - 1)/2 - 2n = 2n(2n-1)(2n-2) / 6 - 2n(2n - 1)/2
2n^2 - n - 2n = 2n(2n -1) [ (2n-2)/6 - 1/2]
12n^2 - 18n = (4n^2 - 2n)(2n - 5)
12n^2 - 18n = 8n^3 - 20n^2 - 4n^2 + 10n
8n^3 - 36n^2 + 28n = 0

So, divide both sides by 4n (because n=/= 0), we have
2n^2 - 9n + 7 = 0

(1+x)^2n
=1+2nx+[2n(2n-1)/2]x^2+[2n(2n-1)(2n-2)/6]x^3+......
∵The coeff. of the 2nd, 3rd, and 4th are in AP
∴2nd term+4th term=2x(3rd term)
∴2n+2n(2n-1)(2n-2)/6=2n(2n-1)
2n[(2n-1)(2n-2)/6+1]=2n(2n-1)
(2n-1)(2n-2)/6+1=2n-1
4n^2-6n+2=12n-12
4n^2-18n+14=0
2n^2-9n+7=0

(1+x)2n=1+ 2nP1 x + 2nP2 x2 + 2nP3 x3 + 2nP4 x4 + ……..

2nP1=(2n)!/(2n-1)! =2n
2nP2=(2n)!/(2n-2)!/2! =n(2n-1)
2nP3=(2n)!/(2n-3)!/3! =n(2n-1)(2n-2)/3= 2n(2n-1)(n-1)/3
2nP4=(2n)!/(2n-4)!/4! =n(2n-1)(2n-2)(2n-3)/12= n(2n-1)(n-1)(2n-3)/6



The 2nd, 3rd and 4th term are in AP, so their different are the same

n(2n-1)(n-1)(2n-3)/6- 2n(2n-1)(n-1)/3= 2n(2n-1)(n-1)/3- n(2n-1)

LHS:
n(2n-1)(n-1)(2n-3)/6- 2n(2n-1)(n-1)/3
= n(2n-1)(n-1)[(2n-3)/6-2/3]
= n(2n-1)(n-1)[(2n-3)-4]/6
= n(2n-1)(n-1)(2n-7)/6

RHS:
2n(2n-1)(n-1)/3- n(2n-1)
= n(2n-1)[(2n-2)/3-1]
= n(2n-1)[(2n-2)-3]/3
= n(2n-1)(2n-5)/3



LHS=RHS
n(2n-1)(n-1)(2n-7)/6= n(2n-1)(2n-5)/3

For n is not zero, and n>1

(n-1)(2n-7)=2(2n-5)
2n2-9n+7=4n-10
2n2-13n+17=0