(1+x+x2)n=a0+a1x+a2x2+⋯+a2nx2n=∑2nr=0arxr
Replace x by 1x.
(1+1x+1x2)2=a0+a1(1x)+a2(1x2)+⋯
or (1+x+x2)nx2n=a0+a1(1x)+⋯+ar(1xr)+⋯
or (1+x+x2)n=a0x2n+a1x2n−1+⋯+arx2n−r+⋯
∑2nr=0arxr=∑2nr=0a2n+rr, by (1)(2)
Equating the coefficients of x2n−r in both sides, we get a2n−r=arorar=a2n−r, where 0 < r < 2n