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