Question

这几道裂项的题目怎么做？

急啊啊啊啊

Answer 1

(1)
an
= [n(n+1) -1 ]/[n(n+1)]
= 1 - 1/[n(n+1)]
= 1 - [1/n- 1/(n+1) ]
Sn
= a1+a2+...+an
=n - [ 1 - 1/(n+1) ]
1/2+5/6 +11/12+19/20+...+9899/9900
=S99
=99 -[ 1 - 1/100]
=98 + 1/100
=9801/100
(2)

an
= 1/ [ (2n)^2 -1]
=(1/2) [1/(2n-1) - 1/(2n+1) ]
Sn
=a1+a2+...+an
=(1/2) [ 1 - 1/(2n+1) ]
1/(2^2-1) +1/4^2-1)+...+1/(100^2-1)
=S50
=(1/2) [ 1 - 1/(100+1) ]
=(1/2) ( 100/101)
=50/101

追问

能帮我做一下最后一道题吗？

追答

an
= 1/[n(n+1)(n+2)]
=(1/2) { 1/[n(n+1)] -1/[(n+1)(n+2) ] }
Sn
=a1+a2+...+an
=(1/2) { 1/2 -1/[(n+1)(n+2) ] }
1/(1x2x3) +1/(2x3x4)+...+1/(99x100x101)
=S99
=(1/2) ( 1/2 -1/10100 )
=(1/2) ( 5049/10100)
=5049/20200

Answer 2

支持一下现形