∑k=0n(kr)(n−ks)=(nr+s)
we know, (1+x)r+s=(1+x)r(1+x)s
∑k=0r+s(nr+s)xn=∑j=0r(jr)xj+∑a=os(as)xa
Now we expand each of the summation on RHS upto x2 -
=[(0r)+(1r)x+(2r)x2+....][(0s)+(1s)x+(2s)x2+...........]=[(0r)(0s)]+[(0r)(1s)+[(0s)(1r)]x+[(1r)(1s)+(0r)(2s)+(2r)(0s)]x2.......
Thus By inspecting a clear Pattern is evident and the n^{th} Power of ''x'' is of the form -
∑k=0n(kr)(n−ks)xk
and This is equated to (kn+m)xk.
Hence, ∑k=0n(kr)(n−ks)=(nr+s)
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