CM: S=$\frac{1}{2^{2}}$+$\frac{1}{3^{2}}$+$\frac{1}{4^{2}}$+…+ $\frac{1}{2020^{2}}$<1

Question

CM:
S=$\frac{1}{2^{2}}$+$\frac{1}{3^{2}}$+$\frac{1}{4^{2}}$+…+ $\frac{1}{2020^{2}}$<1

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Parker 3 tháng 2021-09-11T05:18:21+00:00 1 Answers 2 views 0

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    2021-09-11T05:20:20+00:00

    Đáp án:+Giải thích các bước giải:

    S=$\frac{1}{2^{2}}$+$\frac{1}{3^{2}}$+$\frac{1}{4^{2}}$+…+ $\frac{1}{2020^{2}}$

    Ta có:

    + $\frac{1}{2^{2}}$<$\frac{1}{1.2}$

    + $\frac{1}{3^{2}}$<$\frac{1}{2.3}$

    + $\frac{1}{4^{2}}$<$\frac{1}{3.4}$

    + …………………….

    +$\frac{1}{2020^{2}}$<$\frac{1}{2019.2020}$

    S<$\frac{1}{1.2}$+$\frac{1}{2.3}$+$\frac{1}{3.4}$+…+$\frac{1}{2019.2020}$<1

    S<1-$\frac{1}{2}$+$\frac{1}{2}$-$\frac{1}{3}$+$\frac{1}{3}$-$\frac{1}{4}$+…+$\frac{1}{2019}$-$\frac{1}{2020}$

    S<1-$\frac{1}{2020}$<1

    S<$\frac{2019}{2020}$<1

    =>S<1 (ĐPCM)

    Xin ctrl hay nhất 

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