Question

If the 3rd, 4th , 5th and 6th term in the expansion of `(x+alpha)^n` be, respectively, `a ,b ,ca n dd ,` prove that `(b^2-a c)/(c^2-b d)=(5a)/(3c)dot`

Answer 1

let `t_3 = a`
`t_4 = b`
`t_5 = c`
`t_6 = d`
for the expansion `(x+ alpha)^n`
`t_(r+1)/t_r= (n-r+1)/r* alpha/x`
`t_4/t_3= (n-3+1)/3* alpha/x = b/a = (n-2)/3* alpha/x`
`t_5/t_4= (n-4+1)/4* alpha/x = c/b = (n-3)/4* alpha/x `
`t_6/t_5 = (n-4)/5* alpha/x = alpha/c`
now, `((b^2 - ac)/(bc))/((c^2-bd)/(bc)) = (5a)/(3c) `
`(b^2/(bc) - (ac)/(bc))/(c^2/(bc) - (bd)/(bc)) = (5a)/(3c)`
`(b/c - a/b)/(c/b - d/c) = (5a)/(3c) `
LHS `((4x)/((n-3)*alpha) - (3x)/((n-2)* alpha))/(((n-3)*alpha)/(4*x) - ((n-4)* alpha)/(5x))`
`(x/alpha)/(alpha/x)*((4/(n-3) - 3/(n-2))/((n-3)/4 - (n-4)/5))`
`x^2/alpha^2* (4*5*(4n-8-3n+9))/((n-3)(n-2)(5n-15-4n+16)`
`x^2/ alpha^2 * (4*5)/((n-2)(n-3))`
`(5a)/(3c) = 5/3 * a/b* b/c `
`= 5/3 * (3x)/((n-2)alpha) * (4x)/((n-3)*alpha)`
`= (5*4*x^2)/(alpha^2 (n-2)(n-3))`
`:. LHS = RHS `