Tìm `x` biết `1/21+1/77+1/165+…+1/(n^2+4n)=56/673`.

Tìm `x` biết `1/21+1/77+1/165+…+1/(n^2+4n)=56/673`.

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  1. ta có: $\frac{1}{21}$ + $\frac{1}{77}$ + $\frac{1}{165}$ +…+ $\frac{1}{n²+4n}$ + $\frac{56}{673}$ 

    ⇔ $\frac{1}{3.7}$ + $\frac{1}{7.11}$ + $\frac{1}{11.5}$ +…+ $\frac{1}{n(n+4)}$ + $\frac{56}{673}$ 

    ⇔ $\frac{4}{3.7}$ + $\frac{4}{7.11}$ + $\frac{4}{11.5}$ +…+ $\frac{4}{n(n+4)}$ + $\frac{4.56}{673}$ 

    ⇔ $\frac{1}{3}$ – $\frac{1}{7}$ + $\frac{1}{7}$ – $\frac{1}{11}$ +…+ $\frac{1}{n}$ = $\frac{1}{n+4}$ = $\frac{224}{673}$ 

    ⇔ $\frac{1}{3}$ –  $\frac{1}{n+4}$ = $\frac{224}{673}$ 

    ⇔ $\frac{1}{n+4}$= $\frac{1}{2019}$ 

    ⇔ n = 2015

    ????#ɷįᵰƫ_ᵭậᵱ_ɕɧᶏɨ ????

     

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  2. `1/21 + 1/77 + 1/165 + … + 1/(n^2+4n) = 56/673`

    `⇒ 1/3.7 + 1/7.11 + 1/11.15 + … + 1/(n(n+4)) = 56/673`

    `⇒ 1/3 – 1/7 + 1/7 – 1/11 + 1/11 – 1/15 + … + 1/n – 1/(n+4) = 56/673 xx 4`

    `⇒ 1/3 – 1/(n+4) = 224/673`

    `⇒ 1/(n+4) = 1/3 – 224/673`

    `⇒ 1/(n+4) = 1/2019`

    `⇒ n + 4 = 2019`

    `⇒ n = 2015`

    Vậy `n = 2015`

     

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