ar=an-r
it helps or not [7]
Consider (1-x3)n =(1+x+x2)n (1-x)n = Σ(-1)r nCrx3r
The coefficient of xk on the right hand side is bk = aonCk - a1nCk-1 + ....+(-1)naknC0
If n is a multiple of 3, then let n = 3p
The coefficient of xn on LHS is therefore (-1)p nCp and on the RHS is
bn = aonCn - a1nCn-1 + ....+(-1)nannC0
which is equivalent to the given expression remembering that nCr = nCn-r
Hence the given expression equals (-1)p nCp where p = n/3
phew.. good work prophet!!
Great work in fact...
Cant give 2 pinks for the same post.. or i wud have :)