D=5/4 + 5/4^2 +5/4^3 + 5/4^4 +.....+5/4^99 .Chứng tỏ D<5/3
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DN
0
P
1
13 tháng 8 2016
\(A=4+4^2+4^3+...+4^{100}\)
\(A=\left(4+\text{ }4^2\right)+\left(4^3+4^4\right)+...+\left(4^{99}+4^{100}\right)\)
\(A=\left(1+4\right).\left(4\right)+\left(1+4\right).\left(4^3\right)+...+\left(1+4\right).\left(4^{99}\right)\)
\(A=5.\left(4+4^3+4^5+...+4^{99}\right)\)
Vậy A chia hết cho 5
Các bạn nha!
LT
1
11 tháng 10 2015
5+5^2 +5^3 +5^4...+5^99+5^100
= ( 5+5^2)+(5^3+5^4)+....+(5^99+5^100)
= 5(1+5)+5^3(1+5)+....+5^99(1+5)
= 5.6+5^3.6+....+5^99.6
= (5+5^3+....+5^99).6
Vì (5+5^3+....+5^99).6 chia hết cho 6 nên 5+5^2 +5^3 +5^4...+5^99+5^100 chia hết cho 6.
9 tháng 7 2017
1.
Ta có:
1/2 < 2/3
3/4 < 4/5
.............
99/100 < 100/101
=> 1/2*3/4*5/6*...*99/100 < 2/3*4/5*6/7*...*100/101
=> A < B
2.
\(A\cdot B=\left[\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right]\cdot\left[\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\right]\)
\(A\cdot B=\frac{\left[1\cdot3\cdot5\cdot7\cdot...\cdot99\right]\left[2\cdot4\cdot6\cdot8\cdot...\cdot100\right]}{\left[2\cdot4\cdot6\cdot8\cdot...\cdot100\right]\left[3\cdot5\cdot7\cdot9\cdot...\cdot101\right]}=\frac{1\cdot3\cdot5\cdot...\cdot99}{3\cdot5\cdot7\cdot...\cdot101}=\frac{1}{101}\)
3.
Vì A < B => A.A < A.B => A2 < 1/101 < 1/100
Mà A2 < 1/100 <=> A2 < \(\frac{1}{10}^2\)=> A < 1/10
E
1
29 tháng 10 2014
5^2+5^3+5^4+...+5^98+5^99=(5^2+5^3)+(5^4+5^5)+...+(5^98+5^99)=5^2.(1+5)+5^4.(1+5)+...+5^98.(1+5)=5^2.6+5^4.6+...+5^98.6=6.(5^2+5^4+...+5^98)=5^2+5^4+...+5^98 chia hết cho 6
BN
2
C
18 tháng 12 2018
A = 4 + 42 + 43 + 44 + ... + 499 + 4100
A = ( 4 + 42 ) + ( 43 + 44 ) + ... + (499 + 4100)
A = ( 4 + 42 ) + 43(4 + 42 ) + .... + 499(4 + 42)
A = 20 + 43.20 + .... + 499.20
A = 20 ( 1 + 43 + .... + 499 )
A = 4.5.(1 + 43 + ... + 499 ) ⋮ 5 ( đpcm )
30 tháng 8 2020
a) Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Leftrightarrow2\cdot A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Leftrightarrow2\cdot A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{100}}\)
NM
1
D=5/4+5/4^2+5/4^3+....+5/4^99
4D=4(5/4+5/4^2+5/4^3+....+5/4^99)
4D=5+5/4+5/4^2+5/4^3+....+5/4^98
Lấy 4D-D=(5+5/4+5/4^2+5/4^3+....+5/4^98)-(5/4+5/4^2+5/4^3+....+5/4^99)
=>3D=5-5/4^99
D=5/3-5/3x4^99<5/3
=>d<5/3(ĐPCM)
`Answer:`
\(D=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+\frac{5}{4^4}+...+\frac{5}{4^{99}}\)
\(\Rightarrow4D=5+\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{98}}\)
\(\Rightarrow4D-D=\left(5+\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{98}}\right)-\left(5+\frac{5}{4^2}+\frac{5}{4^3}+\frac{5}{4^4}+...+\frac{5}{4^{99}}\right)\)
\(\Rightarrow3D=5-\frac{5}{4^{99}}\)
\(\Rightarrow D=\left(5-\frac{5}{4^{99}}\right):3\)
\(\Rightarrow D=\frac{5}{3}-\frac{5}{4^{99}.3}< \frac{5}{3}\)
Vậy `D<5/3`