(x + 1) x f(x + 1) + x x f(x) + (x — 1) x f(x — 1)= 0 ...(i) In the above equation, replacing x by x - 1 , we get x × f(x) + ( x - 1) × f ( x - 1) + (x -2 ) × f(x -2 ) = 0 ...(ii) From equations (i) and (ii), we get (x+1)×f(x+1)=(x−2)×f(x−2) ...(iii) Replacing x by x + 2 in equation (iii), we get x×f(x)=(x+3)×f(x+3)=(x+6)×f(x+6) and so on... Hence, f(1)=4f(4)=7f(7)...,2f(2)=5f(5)=8f(8)... and 3f(3)=6f(6)=9f(9)... Also, 3f(3)=6f(6)⇒f(3)=2f(6)⇒f(3)=180×2=360 By putting the values of f(1) and f(3) in f(1)+2f(2)+3f(3)=0, we get f(2) = -560 Also, 2f(2)=14f(14)⇒f(14)=