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b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)
\(B\text{=}2.63+...+2^{56}.63\)
\(\Rightarrow B⋮63\)
\(\Rightarrow B⋮21\)
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 :
\(\dfrac{1}{1^2}=1\)
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
............................
\(\dfrac{1}{50^2}< \dfrac{1}{49.50}\)
\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....................+\dfrac{1}{49.50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...........+\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{50}\)
\(\Rightarrow A< 2-\dfrac{1}{50}< 2\)
\(\Rightarrow A< 2\rightarrowđpcm\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{20^2}\) . CMR : A<1
Giải:
Có \(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\\ \dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\\ ....\\ \dfrac{1}{20^2}< \dfrac{1}{19\cdot20}\)
Nên `A=1/2^2+1/3^2+1/4^2+...+1/(20^2)<1/1.2+1/2.3+1/3.4+...+1/19.20`
`=1-1/2+1/2-1/3+1/3-1/4+...+1/19-1/20=1-1/20=19/20`
Mà `19/20<1`
nên `A<1(đpcm)`
Bài 1:
a) Ta có:
\(3^{200}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=\left(2^3\right)^{100}=8^{100}\)
Vì \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)
b) Ta có:
\(71^{50}=\left(71^2\right)^{25}=5041^{25}\)
\(37^{75}=\left(37^3\right)^{25}=50653^{25}\)
Vì \(5041^{25}< 50653^{25}\Rightarrow71^{50}< 37^{75}\)
c) Ta có:
\(\frac{201201}{202202}=\frac{201.1001}{202.1001}=\frac{201}{202}\)
\(\frac{201201201}{202202202}=\frac{201.1001001}{202.1001001}=\frac{201}{202}\)
\(\Rightarrow\frac{201201}{202202}=\frac{201201201}{202202202}\)
Bài 2:
a) \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}\)
Ta có: \(\frac{1}{1^2}=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(\Rightarrow A< 1+1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A< 1+1-\frac{1}{50}\)
\(\Rightarrow A< 2-\frac{1}{50}< 2\)
b) \(B=2^1+2^2+2^3+...+2^{30}\) (Có 30 số hạng)
\(\Rightarrow B=\left(2^1+2^2+...+2^5+2^6\right)+\left(2^7+2^8+2^9+...+2^{12}\right)+...+\left(2^{25}+2^{26}+...+2^{29}+2^{30}\right)\)
(có \(30:6=5\) nhóm)
\(\Rightarrow B=1\left(2^1+2^2+...+2^6\right)+2^6\left(2^1+2^2+...+2^6\right)+.....+2^{24}\left(2^1+2^2+...+2^6\right)\)
\(\Rightarrow B=1.126+2^6.126+2^{12}.126+...+2^{24}.126\)
\(\Rightarrow B=126.\left(1+2^6+2^{12}+...+2^{24}\right)\)
\(\Rightarrow B=21.6.\left(1+2^6+2^{12}+...+2^{24}\right)⋮21\)
\(\Rightarrow B⋮21\)
Đặt A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
A=\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}\)
Ta thấy :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\)
\(\dfrac{1}{100.100}< \dfrac{1}{99.100}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
Nhân xét :
\(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};\)
\(...;\dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}+...+\)
\(\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A< 1-\dfrac{1}{100}\)
\(\Rightarrow A< \dfrac{99}{100}\)
Vì \(A< \dfrac{99}{100}< 1\)
\(\Rightarrow A< 1\)
Bài 1)
Đặt \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\)
Ta thấy:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4};....;\dfrac{1}{100^2}=\dfrac{1}{100.100}< \dfrac{1}{99.100}\)\(\Rightarrow\) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\) < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)
\(\Rightarrow\) A < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow\) A < \(1-\dfrac{1}{100}\) < 1 \(\Rightarrow\) A < 1
Vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\)< 1
Giải
Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)
\(\Rightarrow\)D < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}\)
Nhận xét: \(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{19.20}=\dfrac{1}{19}-\dfrac{1}{20}\)
\(\Rightarrow\) D< 1- \(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
D< 1 - \(\dfrac{1}{20}\)
D< \(\dfrac{19}{20}\)<1
\(\Rightarrow\)D< 1
Vậy D=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{5^2}\)<1
A=\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
A=\(\dfrac{1}{2^2.1}+\dfrac{1}{2^2.2^2}+\dfrac{1}{3^2.2^2}+...+\dfrac{1}{50^2.2^2}\)
A=\(\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\right)\)
Ta có :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)Nhận xét :
\(\dfrac{1}{1.2}< 1-\dfrac{1}{2};\dfrac{1}{2.3}< \dfrac{1}{2}-\dfrac{1}{3};...;\dfrac{1}{49.50}< \dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{2^2}\left(1-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{4}.\dfrac{49}{50}\)<1
A<\(\dfrac{49}{200}< \dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}\)
1)Ta thấy: \(\dfrac{1}{n^2}=\dfrac{1}{n.n}< \dfrac{1}{\left(n-1\right)n}\)
=>A=\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}...+\dfrac{1}{50^2}< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49.50}\)
A<\(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}=2-\dfrac{1}{50}< 2\)
Vậy A<2
2)Ta có:2S=6+3+\(\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^8}\)
2S-S=(6+3+\(\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^8}\))-(3+\(\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^9}\))
=>S=6-\(\dfrac{3}{2^9}=\dfrac{6.2^9-3}{2^9}\)
Vậy S=\(\dfrac{6.2^9-3}{2^9}\)
\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}=\dfrac{1}{1.1}+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\)\(A=\dfrac{1}{1.1}+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}< \dfrac{1}{1.1}+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\left(1\right)\)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:
\(B=\left(2^1+2^2+2^3+2^4+2^5+2^6\right)+\left(2^7+2^8+2^9+2^{10}+2^{11}+2^{12}\right)+...+\left(2^{25}+2^{26}+2^{27}+2^{28}+2^{29}+2^{30}\right)\)
\(B=2^1\left(1+2+2^2+2^3+2^4+2^5\right)+2^7\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{25}\left(1+2+2^2+2^3+2^4+2^5\right)\)
\(B=2^1.63+2^7.63+...+2^{25}.63\)
\(B=63.\left(2^1+2^7+...+2^{25}\right)⋮63\)
\(B=21.3.\left(2^1+2^7+...+2^{25}\right)⋮21\left(đpcm\right)\)