Bài 4 câu a vài câu B bài 3 ạ 2 câu thôi nhanh hộ mình ạ

thanh352000

2 năm trước

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Câu `3b :`

`A = - 1 + 2 - 2^2 + 2^3 - .... + 2^99`

`2A = - 2 + 2^2 - 2^3 + 2^4 - .... + 2^100`

`2A + A = ( - 1 + 2 - 2^2 + 2^3 - .... + 2^99 ) + ( - 2 + 2^2 - 2^3 + 2^4 - .... + 2^100 )`

`3A = - 1 + 2^100`

`3A = 2^100 - 1`

`A = ( 2^100 - 1 )/3`

Câu `4a :`

Ta có `: A = 1 + 2 + 2^2 + 2^3 + .... + 2^95`

`A = ( 1 + 2 + 2^2 ) + ( 2^3 + 2^4 + 2^5 ) + .... + ( 2^93 + 2^94 + 2^95 )`

`A = 1 . ( 1 + 2 + 2^2 ) + 2^3 . ( 1 + 2 + 2^2 ) + .... + 2^93 . ( 1 + 2 + 2^2 )`

`A = ( 1 + 2 + 2^2 ) . ( 1 + 2^3 + .... + 2^93 )`

`A = 7 . ( 1 + 2^3 + .... + 2^93 ) vdots 7 ( 1 )`

Lại có `: A = ( 1 + 2 ) + ( 2^2 + 2^3 ) + .... + ( 2^94 + 2^95 )`

`A = 1 . ( 1 + 2 ) + 2^2 . ( 1 + 2 ) + .... + 2^94 . ( 1 + 2 )`

`A = ( 1 + 2 ) . ( 1 + 2^2 + .... + 2^94 )`

`A = 3 . ( 1 + 2^2 + .... + 2^94 ) vdots 3 ( 2 )`

Từ `( 1 )` và `( 2 ) ,` ta có `: A vdots 3` và `7 (` Điều phải chứng minh `)`

Vote (0) Phản hồi (0) 2 năm trước

phuongct1981

Đáp án + Giải thích các bước giải:

`A=-1+2-2^2+2^3-...+2^99`

`=>2A=-2+2^2-2^3+2^4-...+2^100`

`=>2A+A=(-2+2^2-2^3+2^4-...+2^100)-(-1+2-2^2+2^3-...+2^99)`

`=>3A=-1+2^100`

`=>A=(2^100-1)/3`

`---`

`A=1+2+2^2+2^3+...+2^95`

`=>A=(1+2)+2^2(1+2)+...+2^94(1+2)`

`=>A=3+2^2 .3+...+2^94 .3`

`=>A=3(1+2^2+...+2^94)\vdots3(1)`

`A=1+2+2^2+2^3+...+2^95`

`=>A=(1+2+2^2)+...+2^93(1+2+2^2)`

`=>A=7+...+2^93 .7`

`=>A=7(1+...+2^93)\vdots7(2)`

Từ `(1),(2)`

Vậy `A\vdots3` và `A\vdots7`

Vote (0) Phản hồi (0) 2 năm trước

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