初一数学题 计算1/3+1/8+1/15+........+1/ n（n+2）=?

Question

如题所述

Answer 1

1/(n*(n+2))=(1/2)*((1/n)-1/(n+2));
所以，原式=（1/2)*[(1/1+1/2+1/3+1/4...+1/n)-(1/3+1/4+1/5+...+1/(n+2))]
=(1/2)*[1+1/2-1/(n+1)-1/(n+2)]本回答被提问者采纳

Answer 2

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

1/3+1/8+1/15+........+1/ n（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))=3/2-1/(2(n+1)(n+2))

Answer 3

1/3+1/8+1/15+........+1/ n（n+2）
=1/1*3+1/2*4+1/3*5+1/4*6+……+1/(n-1)(n+1)+1/n(n+2)
=1/2[1/1-1/3+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/(n+1)+1/2-1/(n+2)]
=(3n^2+5n)/(4n^2+12n+8)

Answer 4

1/ n（n+2）=[1/n-1/(n+2)]/2
根据数列知识解答，可错位想减

Answer 5

裂项
1/ n（n+2）=（1/ n-/ （n+2））/2
再逐个相加可以抵消