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ta có
\(4A=4+4^2+4^3+..+4^{99}+4^{100}=\left(1+4+4^2+..+4^{99}\right)+4^{100}-1\)
hay
\(4A=A+4^{100}-1\Leftrightarrow A=\frac{4^{100}-1}{3}< \frac{4^{100}}{3}=\frac{B}{3}\)
vậy ta có điều phải chứng minh
A=\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+\(\frac{1}{5^2}\)+...+\(\frac{1}{98^2}\)
A=\(\frac{1}{3.3}\)+\(\frac{1}{4.4}\)+\(\frac{1}{5.5}\)+...+\(\frac{1}{98.98}\)
A<\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{97.98}\)=\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+...+\(\frac{1}{97}\)-\(\frac{1}{98}\)=\(\frac{1}{2}\)-\(\frac{1}{98}\)=\(\frac{24}{49}\)<1.
Vậy A<1
1. 3A = 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101
=> 3A - A = (3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 )
=> 2A = 3^101 - 3 => 2A + 3 = 3^101 vậy n = 101
2. 2A = 8 + 2 ^ 3 + 2^4 + ... + 2^20 + 2^21
=> 2A - A = (8 + 2 ^ 3 + 2^4 + ... + 2^20 + 2^21) - (4+ 2^2 + 2 ^ 3 + 2^4 + ... + 2^20 )
=> A = 2^21 là một lũy thừa của 2
3.
a) 3A = 3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101
=> 3A - A = (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (1 + 3 + 3 ^2 + 3 ^ 3 + ... + 3 ^100)
=> 2A = 3^101 - 1 => A = (3^101 - 1)/2
b) 4B = 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101
=> 4B - B = (4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101) - (1 + 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 )
=> 3B = 4^101 - 1 => B = ( 4^101 - 1)/2
c) xem lại đề ý c xem quy luật như thế nào nhé.
d) 3D = 3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151
=> 3D - D = (3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151) - (3 ^100 + 3 ^ 101 + 3 ^ 102 + .... + 3 ^ 150)
=> 2D = 3^ 151 - 3^100 => D = ( 3^ 151 - 3^100)/2
1/1002 + 1/1012 + ... + 1/1992 < 1/99.100 + 1/100.101 + ... + 1/198.199 = 1/99 - 1/100 + 1/100 - 1/101 + ... + 1/198 - 1/199 = 1/99 - 1/199
\(\Rightarrow\)Vậy 1/1002 + 1/1012 + ... + 1/1992 < 1/99 (vì 1/99 đã lớn hơn 1/99 - 1/199 rồi mà G lại còn bé hơn 1/99 - 1/199 nữa)
1/1002 + 1/1012 + ... + 1/1992 > 1/100.101 + ... + 1/199.200 = 1/100 - 1/101 + ... + 1/199 - 1/200 = 1/100 - 1/200 = 1/200
\(\Rightarrow\)Vậy 1/1002 + 1/1012 + ... + 1/1992 > 1/200
Mình ngu lắm dân trần đăng ninh chuyên anh mà làm sao giỏi toán được
Gọi \(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(\forall A>\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
\(\Leftrightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< A< \frac{1}{2}\)
\(\Rightarrowđpcm\)
\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
Vậy
Ta có:
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{99}{100}\)
Mà \(\frac{99}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)
Vậy \(A< 1\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(\frac{1}{4^2}=\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{100^2}=\frac{1}{100\cdot100}< \frac{1}{99\cdot100}\)
=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
=> \(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
=> \(A< \frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)
Lại có : \(\frac{99}{100}< 1\)
=> \(A< \frac{99}{100}< 1\)=> \(A< 1\)( đpcm )
la la
cảm ơn
chứng minh rằng 1/2 mũ 2 +1/3 mũ 2 + 1/4 mũ 2 +.....+1/2018 mũ 2 +1/2019 mũ 2
quá dễ
có chơi lq ko , best flo việt nam
solo flo ko? best flo vn
hay
ồ
ta so sánh từng p/s trong biểu thức
A=122+132+142+...+11002<11.2+12.3+13.4+...+199.100�=122+132+142+...+11002<11.2+12.3+13.4+...+199.100
=11−12+12−13+13−14+...+199−1100=11−12+12−13+13−14+...+199−1100
=11−1100=11−1100
=99100=99100
Mà 99100<199100<1
⇒122+132+142+...+11002<1⇒122+132+142+...+11002<1
Vậy A<1
cho A= 1/3 mũ 2 + 3/3 mũ 4 +5/3 mũ 6 + .....+ 99/3 mũ 100 chứng minh A < 5/32
chứng minh rằng : 1/2 mũ 2 + 1/3 mũ 2 + 1/4 mũ 2 + ... + 1/ 100 mũ 2 < 1
Ta có:
\(A = \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{10 0^{2}} < \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{99.100}\)
\(= \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + . . . + \frac{1}{99} - \frac{1}{100}\)
\(= \frac{1}{1} - \frac{1}{100}\)
\(= \frac{99}{100}\)
Mà \(\frac{99}{100} < 1\)
\(\Rightarrow \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{10 0^{2}} < 1\)
Vậy \(A < 1\)
A=221+321+421+...+10021
Ta có : 122=12⋅2<11⋅2221=2⋅21<1⋅21
132=13⋅3<12⋅3321=3⋅31<2⋅31
142=14⋅4<13⋅4421=4⋅41<3⋅41 ...
11002=1100⋅100<199⋅10010021=100⋅1001<99⋅1001
=> A=122+132+142+...+11002<11⋅2+12⋅3+13⋅4+...+199⋅100A=221+321+421+...+10021<1⋅21+2⋅31+3⋅41+...+99⋅1001
=> A<11−12+12−13+13−14+...+199−1100A<1