Question

[ᴵᴬᴹɴɢᴜʏễɴ●ɴɢọᴄ●ʜùɴɢッ]

\(\text{Bài 4. Chứng tỏ rằng:}\)

\(a\)) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}< 1\)

\(b\)) \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}>1\)

\(c\)) \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)

\(d\)) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}< 1\)

Answer 1

[Ngoc Anh Thai]

a)

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)

b)

 \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)

Answer 2

[Ngoc Anh Thai]

c)

\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)

d) tương tự câu 1