Given P(x)=(1+x+x2)100=a0+a1x+⋯+a200x200
Compute the following sums:
Setting x=1 gives the result for S1: S1=a0+a1+a2+⋯+a200=3100
Next, setting x=−1 yields S−1=a0−a1+a2+⋯+a200=1100=1
Now, we have S2=12⋅(S1+S−1)=3100+12