chứng minh : 1/12 + 1/22 + 1/32 + ... + 1/502 < 173/100
Những câu hỏi liên quan
Bài) Chứng minh rằng
50/51<1+1/22+1/32+1/42+...+1/502<2
M=(100-1).(100-22).(100-32). ... .(100-502)
-> M = (100 – 1).(100 – 2^2). (100 – 3^2)…(100 – 50^2)
M = (100 – 1).(100 – 2^2). (100 – 3^2)… (100 – 9^2) .(100 – 10^2) .(100 – 11^2) …(100 – 50^2)
M = (100 – 1).(100 – 2^2). (100 – 3^2)… (100 – 9^2). (100 – 100) .(100 – 11^2) …(100 – 50^2)
M = (100 – 1).(100 – 2^2). (100 – 3^2)… (100 – 9^2) .0.(100 – 11^2) …(100 – 50^2)
M = 0
Vậy M = 0.
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Cho A=1/12+1/22+1/22+1/32+1/42+..........+ 1/502<2
Chứng minh rằng:
A = 1/3 + 1/32 + 1/33 + ..........+ 1/399 < 1/2
B = 3/12x 22 + 5/22 x 32 + 7/32 x 42 +............+ 19/92 x 102 < 1
C = 1/3 + 2/32 + 3/33 + 4/34 +.........+ 100/3100 ≤ 0
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
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Chứng minh 0<A<1 với A=\(\frac{1}{2}+\frac{1}{^22}+\frac{1}{^32}+....+\frac{1}{^{100}2}\)
1. Chứng minh rằng
A = 2 + 22 + 23 + ... + 2100 chia hết cho 2,3 và 30
2. Chứng minh rằng
B = 3 + 32 + 33 + ... + 32022 chia hết cho 12 và 15
1: \(A=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)
\(=2\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+...+2^{97}\right)\)
\(=30\left(1+2^4+...+2^{96}\right)⋮30\)
2:
\(B=3+3^2+3^3+...+3^{2022}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2021}+3^{2022}\right)\)
\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{2020}\left(3+3^2\right)\)
\(=12\left(1+3^2+...+3^{2020}\right)⋮12\)
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Chứng minh rằng : 1/1^2+1/2^2+1/3^2+...+1/50^2<173/100
Chứng minh
a)B=1/10+1/11+...+1/28 > 1
b)C=1/10.1/11+1/11.1/12+...+1/20.1/21 > 1/20
c)E=1/51+1/52+...+1/100 > 1
d)F=1/2^2+1/3^2+...+1/9^2
Chứng minh 2/5<F<8/9
e)H=1/31+1/32+...+1/60
Chứng minh 3/5<H<4/5
f)K=1/21+1/22+...+1/30>1/3
let S be 1!(12+1+1)+2!(22+2+1)+3!(32+3+1)+...+100!(1002+100+1). Find S+1/101!.(as usual, k! = 1.2.3.....(k-1).k)
Each term of S is n!(n2 + n + 1) = n![n(n + 1) + 1] = n(n + 1)n! + n!
By definition, n(n + 1)n! + n! = n! + n(n + 1)!
Therefore, S can be simplified as
1! + 1.2! + 2! + 2.3! + ... + 100! + 100.101!
So \(\dfrac{S+1}{101!}=\dfrac{1+1!+1\cdot2!+2!+2\cdot3!+...+100!+100\cdot101!}{101!}\)
\(=\dfrac{2!+1\cdot2!+2!+2\cdot3!+3!+...+100!+100\cdot101!}{101!}\)
\(=\dfrac{3!+2\cdot3!+3!+...+100!+100\cdot101!}{101!}\)
\(=\dfrac{4!+3\cdot4!+4!+...+100!+100\cdot101!}{101!}\)
\(=...\)
\(=\dfrac{100!+99\cdot100!+100!+100\cdot101!}{101!}\)
\(=\dfrac{101!+100\cdot101!}{101!}\)
\(=1+100=101\)
Hence, \(\dfrac{S+1}{101!}=101\)
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B = 1/22+1/32+...+1/102
Chứng minh B<1
\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}\)
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(.....\)
\(\dfrac{1}{10^2}< \dfrac{1}{9.10}\)
\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\)
\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}< 1\)
\(\Rightarrow B< 1\left(dpcm\right)\)
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\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}\)
\(B< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{9\times10}\)
\(B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(B< 1-\dfrac{1}{10}\)
\(B< \dfrac{9}{10}< 1\)
Vậy \(B< 1\)
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B= 1/4+1/9+1/16+1/25+1/36+1/49+1/64+1/81+1/100
vì các số kia dần nhỏ dần và số lớn nhất cũng chỉ có 0,25 nên B<1
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