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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)
Ta có :
\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+......................+\dfrac{1}{50^2}\)
Ta thấy :
\(\dfrac{1}{1^2}=1\)
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
............................
\(\dfrac{1}{50^2}< \dfrac{1}{49.50}\)
\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....................+\dfrac{1}{49.50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...........+\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{50}\)
\(\Rightarrow A< 2-\dfrac{1}{50}< 2\)
\(\Rightarrow A< 2\rightarrowđpcm\)
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
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 = \(\dfrac{1}{1^2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+.....+ \(\dfrac{1}{50^2}\)
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)
Ta có:
\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)
\(=1+\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)
Đặt \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\) ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(.....................\)
\(\dfrac{1}{50^2}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
Cộng các vế trên với nhau ta được:
\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(\Rightarrow B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow B< 1-\dfrac{1}{50}< 1\)
\(\Rightarrow1+B< 1+1=2\) Hay \(A< 2\)
Vậy \(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}< 2\) (Đpcm)
\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\\ =1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\\ \Rightarrow A< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}=1+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}=1+1-\dfrac{1}{50}=2-\dfrac{1}{50}< 2\)
Lời giải:
$A< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{99.100}$
$A< \frac{1}{4}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}$
$A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$
$A< \frac{1}{4}+\frac{1}{2}-\frac{1}{100}< \frac{1}{4}+\frac{1}{2}$
Hay $A< \frac{3}{4}$
\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}< 1+\dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+...+\dfrac{1}{50^2-1}\)
\(\Leftrightarrow A< 1+\dfrac{1}{3}+\dfrac{1}{8}+...+\dfrac{1}{2499}\)
\(\Leftrightarrow A< 1+\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+...++\dfrac{1}{48}-\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(\Leftrightarrow A< 1+\dfrac{1}{2}\cdot\left(1-\dfrac{1}{51}+\dfrac{1}{2}-\dfrac{1}{50}\right)\)
\(\Leftrightarrow A< 1+\dfrac{1}{2}\cdot\left(\dfrac{50}{51}+\dfrac{24}{50}\right)\)
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