Question

数列{an},前n项和Sn=n2(n为上标）--4n+1,则Ia1I+Ia2I+``````+Ia10I=?

Answer 1

Sn=n^2-4n+1

Sn-1=(n-1)^2-4(n-1)+1(n>=2)

an=2n-5(n>=2)

a1=-2

a2=-1,

a3=1

下面的项都是正的.

Ia1I+Ia2I+``````+Ia10I

＝a1+a2+a3+...+a10-2(a1+a2)

=-2+(-1)+1+3+,..15-2(-2+(-1))

=-2+9*（16)/2+6

=76