Ta có :

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

Ta thấy :

\(\dfrac{1}{1^2}=1\)

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

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

............................

\(\dfrac{1}{50^2}< \dfrac{1}{49.50}\)

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

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

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

\(\Rightarrow A< 2-\dfrac{1}{50}< 2\)

\(\Rightarrow A< 2\rightarrowđpcm\)

Câu b hướng làm đó là tách con 1/3 và 1/2 ra thành 50 phân số giống nhau. E tách 1/3=50/150 rồi so sánh 1/101, 1/102,...,1/149 với 1/150. Còn vế sau 1/2=50/100 tách tương tự rồi so sánh thôi

2a.

$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}$

$< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{50-49}{49.50}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}$
$=1-\frac{1}{50}< 1$ (đpcm)

2b.

Gọi tổng trên là $T$

Chứng minh vế đầu tiên:

Ta có:

$\frac{1}{101}> \frac{1}{150}$

$\frac{1}{102}> \frac{1}{150}$

....

$\frac{1}{149}> \frac{1}{150}$

$\Rightarrow T> \underbrace{\frac{1}{150}+\frac{1}{150}+...+\frac{1}{150}}_{50}=\frac{50}{150}=\frac{1}{3}$ (đpcm)

Chứng minh vế số 2:

$\frac{1}{101}< \frac{1}{100}$

$\frac{1}{102}< \frac{1}{100}$

....

$\frac{1}{150}< \frac{1}{100}$

$\Rightarrow T< \underbrace{\frac{1}{100}+\frac{1}{100}+....+\frac{1}{100}}_{50}=\frac{50}{100}=\frac{1}{2}$ (đpcm)

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

A = 1 + \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\)+......+\(\dfrac{1}{50.50}\)

      1 = 1

 \(\dfrac{1}{2.2}\)  < \(\dfrac{1}{1.2}\)

  \(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\)

..................

\(\dfrac{1}{50.50}\) < \(\dfrac{1}{49.50}\)

Cộng vế với vế với ta có:

A = \(1+\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\)+....+ \(\dfrac{1}{50.50}\) < 1 + \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+....+\(\dfrac{1}{49.50}\)

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

A < 2 - \(\dfrac{1}{50}\) < 2 ( đpcm)

\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}< 1+\dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+...+\dfrac{1}{50^2-1}\)

\(\Leftrightarrow A< 1+\dfrac{1}{3}+\dfrac{1}{8}+...+\dfrac{1}{2499}\)

\(\Leftrightarrow A< 1+\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+...++\dfrac{1}{48}-\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

\(\Leftrightarrow A< 1+\dfrac{1}{2}\cdot\left(1-\dfrac{1}{51}+\dfrac{1}{2}-\dfrac{1}{50}\right)\)

\(\Leftrightarrow A< 1+\dfrac{1}{2}\cdot\left(\dfrac{50}{51}+\dfrac{24}{50}\right)\)

Nhận xét \(\dfrac{50}{51}< 1;\dfrac{24}{50}< 1\Rightarrow A< 1+\dfrac{1}{2}\cdot\left(\dfrac{50}{51}+\dfrac{24}{50}\right)< 1+\dfrac{1}{2}\cdot\left(1+1\right)=2\)

Vậy A<2

Nhận xét: \(\dfrac{1}{1^2}=1\)

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

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

...........

\(\dfrac{1}{50^2}< \dfrac{1}{49.50}\)

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

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

\(\Rightarrow A< 1+1-\dfrac{1}{50}=2-\dfrac{1}{50}< 2\)

Vậy A < 2

Ta có:

\(\dfrac{1}{1^2}=1;\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{50^2}< \dfrac{1}{40.50}\)

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

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

\(\Rightarrow A< 1+1-\dfrac{1}{50}=2-\dfrac{1}{50}< 2\)

\(\Rightarrow A< 2\)

b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử  \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)

\(B\text{=}2.63+...+2^{56}.63\)

\(\Rightarrow B⋮63\)

\(\Rightarrow B⋮21\)

\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}=\dfrac{1}{1.1}+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\)\(A=\dfrac{1}{1.1}+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}< \dfrac{1}{1.1}+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\left(1\right)\)Mà :\(\dfrac{1}{1}+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}=\dfrac{1}{1}+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=1+1-\dfrac{1}{50}=1+\dfrac{49}{50}=\dfrac{99}{50}< \dfrac{100}{50}=\dfrac{1}{2}\left(2\right)\)

Từ (1) và (2) ta suy ra A<2

B có 30 số hạng, chia B thành 5 nhóm, mỗi nhóm có 6 số hạng như sau:

\(B=\left(2^1+2^2+2^3+2^4+2^5+2^6\right)+\left(2^7+2^8+2^9+2^{10}+2^{11}+2^{12}\right)+...+\left(2^{25}+2^{26}+2^{27}+2^{28}+2^{29}+2^{30}\right)\)

\(B=2^1\left(1+2+2^2+2^3+2^4+2^5\right)+2^7\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{25}\left(1+2+2^2+2^3+2^4+2^5\right)\)

\(B=2^1.63+2^7.63+...+2^{25}.63\)

\(B=63.\left(2^1+2^7+...+2^{25}\right)⋮63\)

\(B=21.3.\left(2^1+2^7+...+2^{25}\right)⋮21\left(đpcm\right)\)

bạn tách số là đc số đẹp