Tìm `x` biết `x+x/(1+2)+x/(1+2+3)+…+x/(1+2+3+…+4041)=4041`.

0 bình luận về “Tìm `x` biết `x+x/(1+2)+x/(1+2+3)+…+x/(1+2+3+…+4041)=4041`.”

1. `x + x/( 1 + 2 ) + x/( 1 + 2 + 3 ) + …. + x/( 1 + 2 + 3 + …. + 4041 ) = 4041`

   `⇔ x . ( 1 + 1/( 1 + 2 ) + 1/( 1 + 2 + 3 ) + …. + 1/( 1 + 2 + 3 + …. + 4041 )) = 4041`

    `⇔ x . ( 1 + 1/3 + 1/6 + …. + 1/8166861 ) = 4041`

   `⇔ x . ( 1 + 2/6 + 2/12 + …. + 2/16333722 ) = 4041`

   `⇔ x . ( 1 + 2/( 2 . 3 ) + 2/( 3 . 4 ) + …. + 2/( 4041 . 4042 )) = 4041`

   `⇔ x . [ 1 + 2 . ( 1/( 2 . 3 ) + 1/( 3 . 4 ) + …. + 1/( 4041 . 4042 )) ] = 4041`

   `⇔ x . [ 1 + 2 . ( 1/2 – 1/3 + 1/3 – 1/4 + …. + 1/4041 – 1/4042 ) ] = 4041`

   `⇔ x . [ 1 + 2 . ( 1/2 – 1/4042 ) ] = 4041`

   `⇔        x . [ 1 + 2 . 1010/2021 ] = 4041`

   `⇔             x . [ 1 + 2020/2021 ] = 4041`

   `⇔                        x . 4041/2021 = 4041`

   `⇔                                     x        = 4041 : 4041/2021`

   `⇔                                     x        = 2021`

   Vậy , `x = 2021 .`

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2. Đáp án:

   `x + x/(1 + 2) +x/(1 + 2 + 3) + … + x/(1 + 2 + 3 + … + 4041) = 4041`

    `⇔ x × [1 + 1/(1 + 2) + 1/(1 + 2 + 3) + … + 1/(1 + 2 + 3 + … + 4041)] = 4041 (1)`

   `text{Xét :}`

   `1/(1 + 2) + 1/(1 + 2 + 3) + … + 1/(1 + 2 + 3 + … + 4041)`

   `= 1/3 + 1/6 + … + 1/8166861`

   `= (1 × 2)/(3 × 2)  + (1×2)/(6×2) + … + (1 × 2)/(8166861×2)`

   `= 2/6 + 2/12 + … + 2/16333722`

   `= 2/(2 × 3) + 2/(3 × 4) + …. + 2/(4041 × 4042)`

   `= 2 × [1/(2 × 3) + 1/(3 × 4) + … + 1/(4041 × 4042)]`

   `= 2 × [1/2 – 1/3 + 1/3 – 1/4 + … + 1/4041 – 1/4042]`

   `= 2 × [1/2 + (- 1/3 + 1/3 – 1/4 + … + 1/4041) – 1/4042]`

   `= 2 × [1/2 – 1/4042]`

   `= 2 × 1010/2021`

   `= 2020/2021`

   `text{Thay vào (1) ta được :}`

   `⇔ x × [1 + 2020/2021] = 4041`

   `⇔ x × [2021/2021 + 2020/2021] = 4041`

   `⇔ x × 4041/2021 = 4041`

   `⇔ x = 4041 ÷ 4041/2021`

   `⇔ x = 2021`

   `text{Vậy x = 2021}`

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