Question

已知an=1/n(n+2)，求Sn

Answer 1

an=1/n(n+2)=(1/2)[1/n-1/(n+2)]
所以，
Sn=(1/2)[1/1-1/3]+(1/2)[1/2-1/4]+(1/2)[1/3-1/5]+。。。+(1/2)[1/(n-1)-1/(n+1)]+(1/2)[1/n-1/(n+2)]
=(1/2)[1-1/3+1/2-1/4+1/3-1/5+。。。+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=(1/2)[1+1/2-1/(n+1)-1/(n+2)] （注：首尾保留的项数相同）
=(1/2)[3(n+1)(n+2)-2(n+2)-2(n+1)]/2(n+1)(n+2)
=(3n²+9n+6-2n-4-2n-2)/4(n+1)(n+2)
=(3n²+5n)/4(n+1)(n+2)

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Answer 2

an=1/n(n+2)
a1=1/1*3=1/2(1-1/3)
a2=1/2*4=1/2(1/2-1/4)
a3=1/3*5=1/2(1/3-1/5)
...
an-1=1/(n-1)(n+1)=1/2(1/(n-1)-1/(n+1))
an=1/n(n+2)=1/2(1/n-1/(n+2))
相加,有:
Sn=1/2(1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+...+1/(n-1)-1/(n+1)+1/n-1/(n+2))
=1/2(1+1/2-1/(n+1)-1/(n+2))
=3/4-(2n+3)/[2(n+1)(n+2)]