$S=$ $\frac{1}{4}$ + $\frac{1}{4^{2}}$ + $\frac{1}{4^{3}}$ + … + $\frac{1}{4^{50}}$

Question

$S=$ $\frac{1}{4}$ + $\frac{1}{4^{2}}$ + $\frac{1}{4^{3}}$ + … + $\frac{1}{4^{50}}$

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Maya 3 tuần 2021-11-21T18:58:33+00:00 2 Answers 4 views 0

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    0
    2021-11-21T19:00:14+00:00

    `S=1/4+1/4^2+1/4^3+…+1/4^50`

    `=> 1/4 S = 1/4(1/4+1/4^2+1/4^3+…+1/4^50)`

    `⇒ 1/4 S = 1/4^2 + 1/4^3 + 1/4^4 +…+ 1/4^51`

    `⇒ S – 1/4 S = (1/4+1/4^2+1/4^3+…+1/4^50)-(1/4^2 + 1/4^3 + 1/4^4 +…+ 1/4^51)`

    `⇒ 3/4 S = 1/4 – 1/4^51`

    `⇒ S = (1/4 – 1/4^51) : 3/4`

    `⇒ S = (1/4 – 1/4^51) . 4/3`

    `⇒ S = 1/3 – 1/(4^50 . 3)`

    0
    2021-11-21T19:00:25+00:00

    $S=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+…+\frac{1}{4^{50}}$

    $⇒4S=\frac{4}{4}+\frac{4}{4^2}+\frac{4}{4^3}+…+\frac{4}{4^{50}}$

    $⇒4S=1+\frac{1}{4}+\frac{1}{4^2}+…+\frac{1}{4^{49}}$

    $⇒4S-S=(1+\frac{1}{4}+\frac{1}{4^2}+…+\frac{1}{4^{49}})-(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+…+\frac{1}{4^{50}})$

    $⇒3S=1-\frac{1}{4^{50}}$

    $⇒S=\frac{1}{3}-\frac{1}{3.4^{50}}$.

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