Question

[Đỗ Phạm Nam Hải]

Chứng minh rằng: M= 1/2² + 1/3² + 1/4² +...+ 1/30² <1

Answer 1

[Pythagoras]

Gọi biểu thức này là \(A\)

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+......+\frac{1}{302^2}\)

Với mọi số tự nhiên ta luôn có : \(\left(n-1\right)^2< n^2< \left(n+1\right)^2\)

\(\Rightarrow\frac{1}{\left(n+1\right)^2}< \frac{1}{n^2}< \frac{1}{\left(n-1\right)^2}\)

Áp dụng ta có :

\(\Rightarrow A< \frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+......+\frac{1}{30^2}\)

\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{29.30}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.........+\frac{1}{29}-\frac{1}{30}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{30}< \frac{1}{2}< 1\)

\(\Rightarrow A< 1\left(đpcm\right)\)

Answer 2

[✎﹏ミ★꧁༺вєѕт↭ℓαυяιєℓ↭νи༻꧂★ミ.༻(Trưở...]

TA CÓ:

\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};.....;\frac{1}{30^2}< \frac{1}{29.30}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{30^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{29.30}\)

ta lại có:\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{29.30}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{30-29}{29.30}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{29}-\frac{1}{30}\)

\(=1-\frac{1}{30}< 1\)

mà \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{30^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{29.30}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{30^2}< 1-\frac{1}{30}< 1\left(đpcm\right)\)