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buitrang1104

3 năm trước

CMR: 1+ $\frac{1}{2^2}$ + $\frac{1}{3^2}$ + $\frac{1}{4^2}$ +...+ $\frac{1}{100^2}$ <2

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Jimsikay09

3 năm trước

Đáp án:

$C < 2$

Giải thích các bước giải:

-Đặt $C = 1 + 1/2^2 + 1/3^2 +........+ 1/100^2$

-Ta có: $1 = 1$

$1/2^2 = 1/1.2$

$1/3^2 = 1/2.3$

................

$1/100^2 < 1/99.100$

⇒ $C < 1 + 1/1.2 + 1/2.3 +........+ 1/99.100$

⇒ $C < 1 + 1 - 1/2 + 1/2 - 1/3 +............+ 1/99 - 1/100$

⇒ $C < 2 - 1/100 < 2$

⇒ $C < 2$

Vậy $C < 2$

nttxyhthkbgd1

3 năm trước

Đáp án:

$\frac{1}{2^2}<\frac{1}{1.2}\\ \frac{1}{3^2}<\frac{1}{2.3}\\ \Rightarrow 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\\ <1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.100}\\ =1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\\ =2-\frac{1}{100}\\ =\frac{199}{100}\\ <\frac{200}{100}=2$

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