Question

已知等差数列{an}中,a2=5,前10项和s10=120

Answer 1

S10=10a1+10*(10-1)*d/2

=10(a2-d)+45d

=10(5-d)+45d

=50+35d

=120

d=2

An=A2+(n-2)d=5+(n-2)*2=2n+1

Sn=na1+n(n-1)*d/2

=n(5-2)+n(n-1)

=n^2+2n

2)

1/Sn=1/[n(n+2)]=(1/2)[1/n-1/(n+2)]

Tn=(1/2)[(1+1/2+1/3+...+1/n)-(1/3+1/4+...+1/n)-1/(n+1)-1/(n+2)]

=(1/2)[1+1/2-1/(n+1)-1/(n+2)]