CMR: 1+ $\frac{1}{2^2}$+ $\frac{1}{3^2}$ + $\frac{1}{4^2}$ +…+ $\frac{1}{100^2}$ <2 13/09/2021 Bởi Adalyn CMR: 1+ $\frac{1}{2^2}$+ $\frac{1}{3^2}$ + $\frac{1}{4^2}$ +…+ $\frac{1}{100^2}$ <2
Đáp án: $\frac{1}{2^2}<\frac{1}{1.2}\\\frac{1}{3^2}<\frac{1}{2.3}\\\Rightarrow 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+…+\frac{1}{10^2}\\<1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+…+\frac{1}{9.100}\\=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+…+\frac{1}{99}-\frac{1}{100}\\=2-\frac{1}{100}\\=\frac{199}{100}\\<\frac{200}{100}=2$ Bình luận
Đáp án:
$\frac{1}{2^2}<\frac{1}{1.2}\\
\frac{1}{3^2}<\frac{1}{2.3}\\
\Rightarrow 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+…+\frac{1}{10^2}\\
<1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+…+\frac{1}{9.100}\\
=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+…+\frac{1}{99}-\frac{1}{100}\\
=2-\frac{1}{100}\\
=\frac{199}{100}\\
<\frac{200}{100}=2$