Question

求和：1/（2²-1）+1/（3²-1）+1/（4²-1）+。。。+1/（n²-1）

Answer 1

1/缓掘（2²-1）+1/禅答（3²-1）+1/（4²-1）+。。。+1/（n²-1）
=1/[（2-1）×（2+1）]+1/[（3-1）×（3+1）]+1/[（4-1）×（4+1）]+1/[（5-1）×（5+1）]。。。贺哪慧+1/[（n-1）×（n+1)]
=1/[1×3]+1/[2×4]+1/[3×5]+1/[4×6]。。。+1/[（n-1）×（n+1)]/2
=[1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+……+1/（n-1)-1/(n+1)]/2
=[1+1/2-1/(n-2)-1/(n+1)]/2

Answer 2

1/（2²-1）+1/（3²-1）+1/（4²-1）+。。。+1/（n²-1）
=1/(2+1)(2-1)+1/(3+1)(3-1)+1/(4+1)(4-1)+…携绝…+1/(n+1)(n-1)
=(1/2)(1-1/辩拍姿3)+(1/2)(1/2-1/4)+(1/2)(1/3-1/贺陵5)+……+(1/2)[1/(n-1)-1/(n+1)]
=(1/2){1+1/2+1/3+1/(n-1)-[1/3+1/4+……+1/(n+1)]}
=(1/2)[1+1/2-1/n-1/(n+1)]

Answer 3

把通式1/（n^-1）拆开凯弯成1/2*（1/(n-1）-1/饥埋(n+1))剩下的你自盯肢闷己搞定，再搞不定那你数学也就太差劲了