1/2mũ2 + 1/4mũ2 + 1/6mũ2+...+1/100mũ2 <1/2
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24 tháng 4 2023
A=1/4(1/1+1/2^2+...+1/50^2)
=>A=1/4+1/4*(1/2^2+...+1/50^2)
=>A<1/4+1/4*(1-1/2+1/2-1/3+...+1/49-1/50)
=>A<1/4+1/4*49/50=99/200<1/2
TB
12 tháng 3 2019
C/m nó nhỏ hơn 3/4 hả bạn ?
Có \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{100}< \frac{3}{4}\)
XO
8 tháng 7 2021
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)(ĐPCM)
3 tháng 7 2019
1)Tính:
a)\(4^2\cdot2=\left(2^2\right)^2\cdot2=2^4\cdot2=2^5=32\)
b)\(36^2:6^2=\left(36:6\right)^2=6^2=48\)
c)\(\left(\frac{2}{5}\right)^{10}:\left(\frac{4}{25}\right)^2=\left(\frac{2}{5}\right)^{10}\cdot\left(\frac{25}{4}\right)^2=\)\(\left(1\right)^{10}\cdot\left(\frac{5}{2}\right)^2=1\cdot\frac{5^2}{2^2}=1\cdot\frac{25}{4}=\frac{25}{4}\)
3 tháng 7 2019
a
\(4^2.2=16.2=32\)
b\(36^2:6^2=36.36:6.6=36.36:36=36\)
c
11 tháng 2 2016
2x(3y-2)+(3y-2) = (2x+1)(3y-2) = -55.Lập bảng :
| 2x+1 | -55 | -11 | -5 | -1 | 1 | 5 | 11 | 55 |
| 3y-2 | 1 | 5 | 11 | 55 | -55 | -11 | -5 | -1 |
| 2x | -56 | -12 | -6 | -2 | 0 | 4 | 10 | 54 |
| 3y | 3 | 7 | 13 | 57 | -53 | -9 | -3 | 1 |
| x | -28 | -6 | -3 | -1 | 0 | 2 | 5 | 27 |
| y | 1 | 19 | -3 | -1 |
Vậy (x;y) = (-28;1);(-1;19);(2;-3);(5;-1)
12 tháng 12 2019
a) Ta có : A=2+22+23+...+210
=(2+22)+(23+24)+...+(29+210)
=2(1+2)+23(1+2)+...+29(1+2)
=2.3+23.3+...+29.3
Vì 3\(⋮\)3 nên 2.3+23.3+...+29.3\(⋮\)3
hay A\(⋮\)3
Vậy A\(⋮\)3.
PM
0
30 tháng 8 2017
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+................+\dfrac{1}{2008^2}\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
...................
\(\dfrac{1}{2008^2}< \dfrac{1`}{2007.2008}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+............+\dfrac{1}{2007.2008}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{2007}-\dfrac{1}{2008}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2008}< 1\)
\(\Leftrightarrow A< 1\rightarrowđpcm\)
NV
3
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
9 tháng 11 2023
D = 40 + 41 + 42 + 43 + 44 + ... + 4200
4.D = 4 + 42 + 43 + 44 + 45 +... + 4201
4D - D = (4 + 42 + 43 + 44 + 45 + ... + 4201) - (40 + 41 + 42 +...+4200)
3D = 4 + 42 + 44 + 44 + 45 + ... + 4201 - 40 - 41 - 42 - ... - 4200
3D = (4 - 41) + (42 - 42) + .... + (4200 - 4200) + 4201 - 40
3D = 4201 - 40
3D + 1 = 4201 - 1 + 1
3D + 1 = 4201
Theo bài ra ta có: 4201 = 4n+1
n + 1 = 201
n = 201 - 1
n = 200
T
9 tháng 11 2023
\(D=4^0+4^1+4^2+4^3+4^4+...+4^{200}\\4D=4\cdot(4^0+4^1+4^2+4^3+4^4+...+4^{200})\\4D=4^1+4^2+4^3+4^4+4^5+...+4^{201}\\4D-D=(4^1+4^2+4^3+4^4+4^5+...+4^{201})-(4^0+4^1+4^2+4^3+4^4+...+4^{200})\\3D=4^{101}-4^0\\3D=4^{101}-1\\\Rightarrow 3D+1=4^{101}\)
Mặt khác: \(3D+1=4^{n+1}\)
\(\Rightarrow 4^{n+1}=4^{101}\\\Rightarrow n+1=101\\\Rightarrow n=101-1=100(tmdk)\)
Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\right)\)
Đặt \(B=1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\)
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
...
\(\dfrac{1}{50^2}< \dfrac{1}{49\cdot50}=\dfrac{1}{49}-\dfrac{1}{50}\)
Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
=>\(B=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 2-\dfrac{1}{50}\)
=>\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{2^2}\left(2-\dfrac{1}{50}\right)=\dfrac{1}{2}-\dfrac{1}{200}< \dfrac{1}{2}\)
\(\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{4}\right)^2+\left(\dfrac{1}{6}\right)^2+...+\left(\dfrac{1}{100}\right)^2\)
\(=\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}.\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}.\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{2}.\dfrac{1}{50}\right)^2\)
\(=\left(\dfrac{1}{2}\right)^2.\left[1+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{50}\right)^2\right]\)
Ta có:
\(\left(\dfrac{1}{2}\right)^2=\dfrac{1}{2.2}< \dfrac{1}{2.1}=\dfrac{2-1}{2.1}=\dfrac{2}{2.1}-\dfrac{1}{2.1}=1-\dfrac{1}{2}\)
\(\left(\dfrac{1}{3}\right)^2=\dfrac{1}{3.3}< \dfrac{1}{3.2}=\dfrac{3-2}{3.2}=\dfrac{3}{3.2}-\dfrac{2}{3.2}=\dfrac{1}{2}-\dfrac{1}{3}\)
...
\(\left(\dfrac{1}{50}\right)^2=\dfrac{1}{50.50}< \dfrac{1}{50.49}=\dfrac{50-49}{50.49}=\dfrac{50}{50.49}-\dfrac{49}{50.49}=\dfrac{1}{49}-\dfrac{1}{50}\)
Khi đó
\(1+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{50}\right)^2< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}=2-\dfrac{1}{50}< 2\)
\(=\left(\dfrac{1}{2}\right)^2.\left[1+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{50}\right)^2\right]< \dfrac{1}{4}.2=\dfrac{1}{2}\)
Vậy \(\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{4}\right)^2+\left(\dfrac{1}{6}\right)^2+...+\left(\dfrac{1}{100}\right)^2< \dfrac{1}{2}\left(đpcm\right)\)
Tick cho mk nha :>>