Chứng minh rằng : 3/1^2×2^2 + 5/2^2×3^2 + 7/3^2×4^2 +….+199/99^2×100^2 < 1 31/08/2021 Bởi Madelyn Chứng minh rằng : 3/1^2×2^2 + 5/2^2×3^2 + 7/3^2×4^2 +….+199/99^2×100^2 < 1
`3/(1^2. 2^2) + 5/(2^2 . 3^2) +…+ 199/(99^2 . 100^2` `= 3/1.4 + 5/4.9+…+ 199/9801.10000` `=1/1 -1/4 + 1/4 -1/9 +…+ 1/9801 – 1/10000` `= 1 – 1/10000 < 1` Vậy `3/(1^2. 2^2) + 5/(2^2 . 3^2) +…+ 199/(99^2 . 100^2` `< 1` Bình luận
`3/(1^2. 2^2) + 5/(2^2 . 3^2) +…+ 199/(99^2 . 100^2`
`= 3/1.4 + 5/4.9+…+ 199/9801.10000`
`=1/1 -1/4 + 1/4 -1/9 +…+ 1/9801 – 1/10000`
`= 1 – 1/10000 < 1`
Vậy `3/(1^2. 2^2) + 5/(2^2 . 3^2) +…+ 199/(99^2 . 100^2` `< 1`