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1 month ago
So, (1+y)n=(y+1)n=∑vk=0n!k!(n−k)!yk1n−k
Hence, the kth term of (1+y)n is Bk=n!k!(n−k)!yk
So, assuming k+2≤n ,
Bk+2=n!(k+2)!(n−k−2)!yk+2=n!k!(n−k)!yk(n−k)(n−k−1)(k+1)(k+2)y2
=Bk(n−k)(n−k−1)(k+1)(k+2)y2