数列{1/[n(n+2)]}中
An=1/[n(n+2)]=(1/2)*2/[n(n+2)]
=(1/2)[(n+2)-n]/[n(n+2)]
=(1/2)[1/n-1/(n+2)]
=(1/2)[1/n-1/(n+1)+1/(n+1)-1/(n+2)]
=(1/2)[P(n)+Q(n)]
P(n)的前n项和Pn=1-1/(n+1)=n/（n+1）
Q(n）的前n项和Qn=1/2-1/（n+2)=n/(2*（n+2))
其前n项和Sn=Pn+Qn
=1/2 ( n/（n+1）+ n/(2*（n+2)))
=1/4 ( 2n/（n+1）+ n/（n+2))
=n(3n+5)/[4(n+1)(n+2)].
数列{1/[n(n+2)]}中
An=1/[n(n+2)]=(1/2)*2/[n(n+2)]
=(1/2)[(n+2)-n]/[n(n+2)]
=(1/2)[1/n-1/(n+2)]
所以和Sn
=(1/2){(1/1-1/3)+(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+……+[1/(n-2)-1/n]+[1/(n-1)-1/(n+1)]+[1/n-1/(n+2)]}
=(1/2)[1+1/2-1/(n+1)-1/(n+2)]
=(1/2){3/2-(2n+3)/[(n+1)(n+2)]}
=n(3n+5)/[4(n+1)(n+2)].
Ak=(1/k)-1/(k+1) Sn=(1-1/2)+(1/2-1/3)+…+[（1/n）-1/（n+1）] =1-[1/（n+1）] =n/（n+1）
答：这是一个等差数列与一个等比数列的对应项的乘积构成的数列的前n项的和，可用用乘等比数列的公比1/2以后作差来解决 Sn=1*(1/2)+2*(1/2^2+3*(1...详情>>
答：详情>>
