Question

a, \(P=\dfrac{1+2}{1^2\cdot2^2}+\dfrac{2+3}{2^2\cdot3^2}+...+\dfrac{9+10}{9^2\cdot10^2}\)

Chứng minh rằng: \(P< 1\)

b, \(Q=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)

Chứng minh rằng: \(Q< \dfrac{1}{2}\)

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Mình cần gấp lắm. 13/08/2017 là mình cần rồi. 

Answer 1

a)  \(P=\frac{1+2}{1^2.2^2}+\frac{2+3}{2^2.3^2}+...+\frac{9+10}{9^2.10^2}\)

\(P=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\) ( rút gọn số mũ nhé )

\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\)

\(P=1-\frac{1}{10}=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)

Vì \(\frac{9}{10}< 1\Rightarrow P< 1\) (đpcm)

b) Chút nữa mình làm nhé ^^

Answer 2

b) 

\(Q=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)

Ta so sánh giữa A và Q.

\(\frac{1}{1.2}>\frac{1}{3};\frac{1}{2.3}>\frac{1}{3^2};\frac{1}{3.4}>\frac{1}{3^3};....;\frac{1}{100.101}>\frac{1}{3^{100}}\)

\(\Rightarrow Q< A\)

Ta lại tiếp tục so sánh A và \(\frac{1}{2}\)

Ta có :

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)

\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)

\(\Rightarrow A=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\Leftrightarrow A< \frac{1}{2}\)

Ta được:

\(Q< A< \frac{1}{2}\Leftrightarrow Q< \frac{1}{2}\)

Answer 3

b) \(Q=\frac{1}{3}\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)\)

\(Q=\frac{1}{3}\left(1+Q-\frac{1}{3^{100}}\right)\)

\(Q=\frac{1}{3}+\frac{1}{3}Q-\frac{1}{3^{101}}\)

\(\frac{2}{3}Q=\frac{1}{3}-\frac{1}{3^{101}}\)

\(\left(=\right)Q=\left(\frac{1}{3}-\frac{1}{3^{101}}\right):\frac{2}{3}=\left(\frac{1}{3}-\frac{1}{3^{101}}\right).\frac{3}{2}\)

\(Q=\frac{1}{2}-\frac{1}{3^{100}.2}< \frac{1}{2}\) (đpcm)