Let d be the common difference of the given A.P.
Then since a1=0, we have a2=d,
a3=2d,....an=(n−1)d.
Hence L.H.S. =2dd+3d2d+4d3d+...+(n−1)d(n−2)d−d(1d+12d+13d+...+1(n−3)d)
=(1+1)+(1+12)+(1+13)+...+(1+1n−3)+(1+1n−2)−(1+12+13+...+1n−3)
=1+1+1+... to (n−2) terms +1n−2
=(n−2)+1n−2=an−1a2+a2an−1.