Question

①已知数列{an}的an=(-1)n次方(2n-1)求数列an前40项和.②求和:sn=1/2×4+1/3×5+1/4×

Answer 1

①an=(-1)^n*(2n-1),

数列an前40项和=（-1+3）+（-5+7）+……+（-77+79）

=2*20=40.

②1/[(n+1)(n+3)]=(1/2)[1/(n+1)-1+3)],

∴Sn=(1/2)[1/2-1/4+1/3-1/5+1/4-1/6+……+1/(n+1)-1/(n+3)]

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

=5/12-(2n+5)/[2(n+2)(n+3)].

③（1）an=1+2+2^2+……+2^(n-1)=2^n-1.

(2)Sn=2^(n+1)-2-n.