Question

Chứng tỏ rằng:

\(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}< 1\)

Thánh nào giúp tui vs tui còn phải làm 3 bài nữa

Answer 1

Nhận xét:

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)

\(..................\)

\(\dfrac{1}{10^2}=\dfrac{1}{10.10}< \dfrac{1}{9.10}\)

Cộng các vế trên với nhau ta được:

\(D< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

\(\Rightarrow D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\Rightarrow D< 1-\dfrac{1}{10}< 1\)

Vậy \(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}< 1\) (Đpcm)

Answer 2

\(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\) \(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}< 1\)

Answer 3

mink làm tắt nha

\(D< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\)

\(D< 1.\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{2}{5}< 1\Rightarrow dpcm\)