Question

Đã trả lời bởi chuyên gia

Answer 1

`a,`

đề sai :

`b,`

Ta có `:`

`1/(2^2) + 1/(3^2) + 1/(4^2) + ... + 1/(n^2) < 1/(1.2) + 1/(2.3) + 1/(3.4) + ... + 1/((n-1) .n)`

`=> 1/(2^2) + 1/(3^2) + 1/(4^2) + ... + 1/(n^2) < 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/(n-1) - 1/n`

`=> 1/(2^2) + 1/(3^2) + 1/(4^2) + ... + 1/(n^2) < 1 - 1/n < 1`

`=> 1/(2^2) + 1/(3^2) + 1/(4^2) + ... + 1/(n^2) <1`

Vậy `1/(2^2) + 1/(3^2) + 1/(4^2) + ... + 1/(n^2) < 1`

Answer 2

`a,`

đề sai :

`b,`

Ta có `:`

`1/(2^2) + 1/(3^2) + 1/(4^2) + ... + 1/(n^2) < 1/(1.2) + 1/(2.3) + 1/(3.4) + ... + 1/((n-1) .n)`

`=> 1/(2^2) + 1/(3^2) + 1/(4^2) + ... + 1/(n^2) < 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/(n-1) - 1/n`

`=> 1/(2^2) + 1/(3^2) + 1/(4^2) + ... + 1/(n^2) < 1 - 1/n < 1`

`=> 1/(2^2) + 1/(3^2) + 1/(4^2) + ... + 1/(n^2) <1`

Vậy `1/(2^2) + 1/(3^2) + 1/(4^2) + ... + 1/(n^2) < 1`

`c,`

Ta có:

`@ A = 2 + 2^2 + 2^3 + ... + 2^60`

`=> A = (2+2^2) + (2^3 + 2^4) + ... + (2^59 + 2^60)`

`=> A = 6 + 2^2 . (2 + 2^2) + ... + 2^58 . (2 + 2^2)`

`=> A = 6 + 2^2 . 6 + ... + 2^58 . 6`

`=> A = 6.(1+2^2+...+2^58)`

Vì `6` $\vdots$ `3 => 6.(1+2^2+...+2^58)` $\vdots$ `3` Hay `A` $\vdots$ `3`

`@ A = 2 + 2^2 + 2^3 + ... + 2^60`

`=> A = (2+2^2 + 2^3) + ... + (2^58 + 2^59 +2^60)`

`=> A = 14 + ... + 2^57 . (2 + 2^2 + 2^3)`

`=> A = 14 + ... + 2^57 . 14`

`=> A = 14 . (1+...+2^57)`

Vì `14` $\vdots$ `7 => 14.(1+...+2^57)` $\vdots$ `7` Hay `A` $\vdots$ `7`

`@ A = 2 + 2^2 + 2^3 + ... + 2^60`

`=> A = (2+2^2+2^3+2^4) + .... + (2^57 + 2^58 + 2^59 + 2^60)`

`=> A = 30 + ... + 2^56 . (2 + 2^2 + 2^3 + 2^4)`

`=> A = 30 + ... + 2^56 . 30`

`=> A = 30. (1+...+2^56)`

Vì `30` $\vdots$ `15 => 30.(1+...+2^56)` $\vdots$ `15` Hay `A` $\vdots$ `15`

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Answer 3

`a,`

đề sai :

`b,`

Ta có `:`

`1/(2^2) + 1/(3^2) + 1/(4^2) + ... + 1/(n^2) < 1/(1.2) + 1/(2.3) + 1/(3.4) + ... + 1/((n-1) .n)`

`=> 1/(2^2) + 1/(3^2) + 1/(4^2) + ... + 1/(n^2) < 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/(n-1) - 1/n`

`=> 1/(2^2) + 1/(3^2) + 1/(4^2) + ... + 1/(n^2) < 1 - 1/n < 1`

`=> 1/(2^2) + 1/(3^2) + 1/(4^2) + ... + 1/(n^2) <1`

Vậy `1/(2^2) + 1/(3^2) + 1/(4^2) + ... + 1/(n^2) < 1`

`c,`

Ta có:

`@ A = 2 + 2^2 + 2^3 + ... + 2^60`

`=> A = (2+2^2) + (2^3 + 2^4) + ... + (2^59 + 2^60)`

`=> A = 6 + 2^2 . (2 + 2^2) + ... + 2^58 . (2 + 2^2)`

`=> A = 6 + 2^2 . 6 + ... + 2^58 . 6`

`=> A = 6.(1+2^2+...+2^58)`

Vì `6` $\vdots$ `3 => 6.(1+2^2+...+2^58)` $\vdots$ `3` Hay `A` $\vdots$ `3`

`@ A = 2 + 2^2 + 2^3 + ... + 2^60`

`=> A = (2+2^2 + 2^3) + ... + (2^58 + 2^59 +2^60)`

`=> A = 14 + ... + 2^57 . (2 + 2^2 + 2^3)`

`=> A = 14 + ... + 2^57 . 14`

`=> A = 14 . (1+...+2^57)`

Vì `14` $\vdots$ `7 => 14.(1+...+2^57)` $\vdots$ `7` Hay `A` $\vdots$ `7`

`@ A = 2 + 2^2 + 2^3 + ... + 2^60`

`=> A = (2+2^2+2^3+2^4) + .... + (2^57 + 2^58 + 2^59 + 2^60)`

`=> A = 30 + ... + 2^56 . (2 + 2^2 + 2^3 + 2^4)`

`=> A = 30 + ... + 2^56 . 30`

`=> A = 30. (1+...+2^56)`

Vì `30` $\vdots$ `15 => 30.(1+...+2^56)` $\vdots$ `15` Hay `A` $\vdots$ `15`

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