If`a_(1),a_(2),a_(3),"........",a_(n)` are in AP with `a_(1)=0`, prove that `(a_(3))/(a_(2))+(a_(4))/(a_(3))+"......"+(a_(n))/(a_(n-1))-a_(2)((1)/(a_(2))+(1)/(a_(3))"+........"+(1)/(a_(n-2)))=(a_(n-1))/(a_(2))+(a_(2))/(a_(n-1))`.

Explanation

`a_(1),a_(2),a_(3),"........",a_(n)` are in AP with `a_(1)=0` and common difference d
`:.a_(2)=d,a_(3)=2d,"......",a_(n)=(n-1)d" " [dne 0]`
`LHS=(a_(3))/(a_(2))+(a_(4))/(a_(3))+(a_(5))/(a_(4))+"......"+(a_(n))/(a_(n-1))-a_(2)((1)/(a_(2))+(1)/(a_(3))"+........"+(1)/(a_(n-2)))`
`=(1)/(a_(2))(a_(3)-a_(2))+(1)/(a_(3))(a_(4)-a_(2))+"......"+((a_(n-1)-a_(2)))/(a_(n-2))+(a_(n))/(a_(n-1))`
`=(1)/(d)(2d-d)+(1)/(2d)(3d-d)+"...."+([(n-2)d-d])/((n-3)d)+((n-1)d)/((n-2)d)`
`=[1+1+"......."+(n-3)" times "]+(n-1)/(n-2)`
`=[1+1+"......."+(n-3)" times "]+(n-1)/(n-2)`
`=(n-3)+1+(1)/(n-2)=n-2+(1)/(n-2)`
`=(a(n-2)d)/(d)+(d)/((n-2)d)=(a_(n-1))/(a_(2))+(a_(2))/(a_(n-1))RHS`.