Ta có:
1/2^2 > 1/2.3
1/3^2 > 1/3.4
...
1/10^2 > 1/10.11
-> Cộng dọc theo vế ta có:
1/2^2+1/3^2+...+1/10^2 > 1/2.3+1/3.4+...+1/10.11
                                         = 1/2-1/3+1/3-1/4+...+1/10-1/11 

                                         = 1/2 - 1/11 = 9/22  (đpcm)         

1/2^2+1/3^2+...+1/50^2<1/1*2+1/2*3*+...+1/49*50

=1/1-1/2+1/2-1/3+...+1/49-1/50<1

=>S<1+1=2

\(\dfrac{1}{3^2}>\dfrac{1}{3\cdot4}=\dfrac{1}{3}-\dfrac{1}{4}\)

\(\dfrac{1}{4^2}>\dfrac{1}{4\cdot5}=\dfrac{1}{4}-\dfrac{1}{5}\)

...

\(\dfrac{1}{100^2}>\dfrac{1}{100}-\dfrac{1}{101}\)

Do đó: \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}>\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}>\dfrac{90.9}{303}=\dfrac{3}{10}\)(1)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3}-\dfrac{1}{4}\)

...

\(\dfrac{1}{100^2}< \dfrac{1}{99}-\dfrac{1}{100}\)

Do đó: \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

=>\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}< \dfrac{50}{100}=\dfrac{1}{2}\)(2)

Từ (1),(2) suy ra \(\dfrac{3}{10}< \dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)

ơi

bé

không

Để chứng minh 3<S<6, ta cần tính giá trị của biểu thức S và thấy xem nó có nằm trong khoảng (3, 6) hay không.

Đầu tiên, ta tính tổng S bằng cách đặt S bên cạnh tổng harmonic thứ 63, rồi trừ đi tổng harmonic thứ 62:

S = 1/1 + 1/2 + 1/3 + ... + 1/63 S - 1/2 = 1/2 + 1/3 + ... + 1/63

Lặp lại phương pháp trên đối với S - 1/2, ta có:

S - 1/2 - 1/3 = 1/3 + ... + 1/63

Cứ lặp lại phương pháp trên đến khi ta được:

S - 1/2 - 1/3 - ... - 1/62 = 1/63

Tổng quát lại, ta có:

S - 1/2 - 1/3 - ... - 1/62 - 1/63 = 0

Từ đây suy ra:

3/2 < 1/2 + 1/3 + ... + 1/62 + 1/63 < 1 + 1/2 + 1/3 + ... + 1/62 < 6

Vì vậy, ta có:

3 < S < 6

Vậy, ta đã chứng minh được rằng 3<S<6.

Bài 2 :

\(S=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+............+\dfrac{2017}{4^{2017}}\)

\(\Leftrightarrow4S=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...........+\dfrac{2017}{4^{2016}}\)

\(\Leftrightarrow4S-S=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+..........+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+..........+\dfrac{2017}{4^{2017}}\right)\)

\(\Leftrightarrow3S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+.........+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2016}}\)

Đặt :

\(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2016}}\)

\(\Leftrightarrow4A=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2015}}\)

\(\Leftrightarrow4A-A=\left(4+1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2016}}\right)\)

\(\Leftrightarrow3A=4-\dfrac{1}{4^{2016}}\)

\(\Leftrightarrow D=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}\)

\(\Leftrightarrow3S=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}-\dfrac{2017}{4^{2016}}\)

\(\Leftrightarrow3S< \dfrac{4}{3}\)

\(\Leftrightarrow S< \dfrac{4}{9}\)

\(\Leftrightarrow S< \dfrac{1}{2}\rightarrowđpcm\)

\(A=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\) ( A cho đẹp :v)

\(4A=4\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)

\(4A=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\)

\(4A-A=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)\(3A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}\)

Đặt:

\(M=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\)

\(4M=4\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)

\(4M=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\)

\(4M-M=\left(4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)\(3M=4-\dfrac{1}{4^{2016}}\)

\(M=\dfrac{4}{3}-\dfrac{1}{4^{2016}}\)

Thay M vào A ta có:

\(A=\dfrac{4}{9}-\dfrac{1}{4^{2016}.3}-\dfrac{2017}{4^{2017}}\)

\(\Rightarrow A< \dfrac{1}{2}\Rightarrowđpcm\)

\(S=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)

Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)

\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

\(A< \dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{50-49}{49.50}\)

\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..+\dfrac{1}{49}-\dfrac{1}{50}\)

\(A< 1-\dfrac{1}{50}\Rightarrow A< 1\)

Ta có \(S=\dfrac{1}{2^2}\left(1+A\right)\)

Ta có

\(A< 1\Rightarrow1+A< 2\Rightarrow S< \dfrac{1}{2^2}.2=\dfrac{1}{2}\)

4S=1+24+342+....+2014420134S=1+24+342+....+201442013

4S−S=3S=1+24+342+....+201442013−(14+242+343+....+201442014)4S−S=3S=1+24+342+....+201442013−(14+242+343+....+201442014)

3S=1+(24−14)+(342−242)+......+(201442013−201342013)−2014420143S=1+(24−14)+(342−242)+......+(201442013−201342013)−201442014

3S=1+14+142+143+.....+142013−2014420143S=1+14+142+143+.....+142013−201442014

đặt A=1+14+142+143+....+142023A=1+14+142+143+....+142023

4A−A=4+1+14+142+.....+142022−(1+14+142+....+142023)4A−A=4+1+14+142+.....+142022−(1+14+142+....+142023)

3A=4−1420233A=4−142023

A=43−13.42023A=43−13.42023

⇒3S=43−13.42023−201442024⇒3S=43−13.42023−201442024

⇒S=49−19.42023−20143.42024⇒S=49−19.42023−20143.42024

do 49<48=1249<48=12

⇒S=49−19.42023−20143.42024<48=12(đpcm)

Dễ quá

ohh

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