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−n2
2
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