Đề bài là chứng minh nhé
Ta có:
`\qquad 1-1/2+1/3-1/4+...+1/{2n-1}-1/{2n}`
`=(1+1/3+1/5+...+1/{2n-1})`
`-(1/2+1/4+...+1/{2n})`
`=(1+1/2+1/3+1/4+...+1/{2n-1}+1/{2n})`
`-(1/2+1/4+...+1/{2n})-(1/2+1/4+...+1/{2n})`
`=(1+1/2+1/3+1/4+...+1/{2n-1}+1/{2n})`
`-2.(1/2+1/4+...+1/{2n})`
`=(1+1/2+1/3+1/4+...+1/n+1/{n+1}+1/{n+2}+...+1/{2n-1}+1/{2n})`
`-(1+1/2+...+1/{n})`
`=1/{n+1}+1/{n+2}+...+1/{2n}`
Vậy: `1-1/2+1/3-1/4+...+1/{2n-1}-1/{2n}`
`=1/{n+1}+1/{n+2}+...+1/{2n}`
