求和:(a-1)+(a2-2)+(a3-3)+…+(an-n).
问题描述:
求和:(a-1)+(a2-2)+(a3-3)+…+(an-n).
答
S=(a-1)+(a2-2)+(a3-3)+…+(an-n)=(a+a2+a3+…+an)-(1+2+3+…+n)当a=0时,S=-(1+2+3+…+n)=-n(n+1)2;当a=1时,S=n−n22;当a≠1,且a≠0时,a+a2+a3+…+an=a(1−an)1−a,∴S=a(1−an)1−a−n(n+1)2...