计算1÷〔x(x+1)〕+1÷〔(x+1)(X+2)〕+1÷〔(x+2)(x+3)〕+.+1÷〔(x+1998)(x+1999)〕并求当x=1时,该代数式的值.为什么1÷〔x(x+1)〕等于1/X-1/（X+1）?
2019-05-07
计算1÷〔x(x+1)〕+1÷〔(x+1)(X+2)〕+1÷〔(x+2)(x+3)〕+.+1÷〔(x+1998)(x+1999)〕
并求当x=1时,该代数式的值.
为什么1÷〔x(x+1)〕等于1/X-1/（X+1）?
优质解答
1/[x(x+1)+1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+...+1/[(x+1998)(x+1999)]
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+1998)-1/(x+1999)
=1/x-1/(x+1999)
当x=1时,1-1/2000=1999/2000
拆项啊~~
1/[x(x+1)]=[(x+1)-x]/[x(x+1)]=1/x-1/(x+1)
1/[(x+1)(x+2)]=[(x+2)-(x+1)]/[(x+1)(x+2)]=1/(x+1)-1/(x+2)
.
.
.
1/[(x+n)(x+n+1)]=[(x+n+1)-(x+n)]/[(x+n)(x+n+1)]=1/(x+n)-1/(x+n+1)
拆项是数列求和的常用方法,要掌握
1/[x(x+1)+1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+...+1/[(x+1998)(x+1999)]
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+1998)-1/(x+1999)
=1/x-1/(x+1999)
当x=1时,1-1/2000=1999/2000
拆项啊~~
1/[x(x+1)]=[(x+1)-x]/[x(x+1)]=1/x-1/(x+1)
1/[(x+1)(x+2)]=[(x+2)-(x+1)]/[(x+1)(x+2)]=1/(x+1)-1/(x+2)
.
.
.
1/[(x+n)(x+n+1)]=[(x+n+1)-(x+n)]/[(x+n)(x+n+1)]=1/(x+n)-1/(x+n+1)
拆项是数列求和的常用方法,要掌握