lim( $\frac{1}{n^{2}}$+ $\frac{2}{n^{2}}$ +…+ $\frac{n-1}{n^2}$ ) 17/10/2021 Bởi Camila lim( $\frac{1}{n^{2}}$+ $\frac{2}{n^{2}}$ +…+ $\frac{n-1}{n^2}$ )
`lim(\frac{1}{n²} + \frac{2}{n²} + …+ \frac{n-1}{n²})` Ta có: `\frac{1}{n²} + \frac{2}{n²} + …+ \frac{n-1}{n²}` `= \ frac{1}{n²} (1+2+…+n-1)` `=\frac{1}{n²}. \frac{(n-1)(1+n-1)}{2}` `= \frac{(n-1)n}{2n²}` `= \frac{n²-n}{2n²}` `=> lim(\frac{1}{n²} + \frac{2}{n²} + …+ \frac{n-1}{n²})` `=lim \frac{n²-n}{2n²}` `=lim \frac{1-\frac{1}{n}}{2}` `= \frac{1-0}{2}` `=\frac{1}{2}` Bình luận
Đáp án:
Giải thích các bước giải:
`lim(\frac{1}{n²} + \frac{2}{n²} + …+ \frac{n-1}{n²})`
Ta có:
`\frac{1}{n²} + \frac{2}{n²} + …+ \frac{n-1}{n²}`
`= \ frac{1}{n²} (1+2+…+n-1)`
`=\frac{1}{n²}. \frac{(n-1)(1+n-1)}{2}`
`= \frac{(n-1)n}{2n²}`
`= \frac{n²-n}{2n²}`
`=> lim(\frac{1}{n²} + \frac{2}{n²} + …+ \frac{n-1}{n²})`
`=lim \frac{n²-n}{2n²}`
`=lim \frac{1-\frac{1}{n}}{2}`
`= \frac{1-0}{2}`
`=\frac{1}{2}`