xn=a0+a1(1+x)+a2(1+x)2+…+an(1+x)n=b0+b1(1−x)+b2(1−x)2+…+bn(1−x)n then for n = 101,a50,b50 equals:
−101C50,101C50
101C50,−101C50
−101C50,−101C50
101C50,101C50
xn=[(1+x)−1]n=[1−(1−x)]n=∑k=0n nCk(1+x)n−k(−1)k=∑k=0n nCk(−1)k(1−x)k
∴ak = coefficient of (1+x)k in
∑k=0n nCk(1+x)n−k(−1)k=(−1)n−knCn−k=(−1)n−knCk
and bk=nCk(−1)k
For n=101,k=51 we get
a51,b51= 101C51,−101C51