A= 1/3-1/3^2+1/3^3-1/3^4+…+1/3^99-1/3^10 2) cho B=2n+2/2n-4(n thuoc Z) tim nthuoc Z de B thuoc Z ai nhanh mik cho 5sao

0 bình luận về “A= 1/3-1/3^2+1/3^3-1/3^4+…+1/3^99-1/3^10 2) cho B=2n+2/2n-4(n thuoc Z) tim nthuoc Z de B thuoc Z ai nhanh mik cho 5sao”

1. Đáp án: 

   Giải thích các bước giải:

    1) `A=  1/3-1/3^2+1/3^3-1/3^4+…+1/3^99-1/3^100`

   `3A=  1 -1/3 + 1/3^2 – 1/3^3 + … + 1/3^98 – 1/3^99`

   `=> 3A+ A= 1/3-1/3^2+1/3^3-1/3^4+…+1/3^99-1/3^100 + 1 -1/3 + 1/3^2 – 1/3^3 + … + 1/3^98 – 1/3^99`

   `=> 4A= 1+ (1/3 – 1/3) + (1/3^2 – 1/3^2) + … + (1/3^99 – 1/3^99 ) – 1/3^100`

   ` 4A= 1 – 1/3^100`

   ` A= (1- 1/3^100) : 4`

   ` A= 1/4 – 1/4.3^100`

   2) `B = {2n+2} / {2n-4}`

   Để `B∈ZZ⇔ 2n+2` chia hết cho `2n – 4`

   `<=> n+1` chia hết cho `n-2`

   `<=> n-2 + 3` chia hết cho `n-2`

   `<=> 3` chia hết cho `n-2`

   `=> n-2∈Ư(3)={1;3;-1;-3}`

   `<=> n∈{3; 5; 1; -1}.`

   Vậy `n∈{3; 5; 1; -1} ` thì `B` nguyên.

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