If 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=201,a101,b101 is equal to:
−201C101,−201C101
201C101,−201C101
−201C101,201C101
201C101,201C101
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=201,k=101,
we get a101,b101= 201C101,−201C101