cho b = 1/2 +(1/2)^2+(1/2)^3+...+(1/2)^98+(1/2)^99
cmr b<1
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ND
1
NN
24 tháng 8 2023
\(A=1+2+2^2+2^3+...+2^{99}\)
\(=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{96}+2^{97}+2^{98}\right)+2^{99}\)
\(=7+2^3\left(1+2+2^2\right)+...+2^{96}\left(1+2+2^2\right)+2^{99}\)
\(=7+2^3.7+...+2^{96}.7+2^{99}\)
\(=7\left(1+2^3+...+2^{96}\right)+2^{99}\)
Vì \(7⋮7=>7\left(1+2^3+...+2^{96}\right)⋮7\) mà \(2^{99}⋮̸7\)
\(=>A⋮̸7\)
M
0
25 tháng 8 2023
a: \(A=\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)
\(=\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+\dfrac{100}{100}\)
\(=100\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)=100B
=>B/A=1/100
b: \(A=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\left(1\right)\)
\(=\dfrac{50}{49}+\dfrac{50}{48}+....+\dfrac{50}{2}+\dfrac{50}{50}\)
\(=50\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(B=\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{49}+\dfrac{2}{50}\)
\(=2\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)
=>A/B=25
CL
1
GR
2
18 tháng 7 2018
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+.....+\left(\frac{1}{2}\right)^{99}\)
\(=\frac{1}{2}+\frac{1}{2^2}+\frac{2}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\)
Ta có : \(\frac{1}{2}< \frac{1}{1};\frac{1}{2^2}< \frac{1}{1\cdot2};.....;\frac{1}{2^{99}}< \frac{1}{98\cdot99}\)
\(\Rightarrow B=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}< 1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{98\cdot99}\)
\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{98\cdot99}=1+1-\frac{1}{99}=2-\frac{1}{99}\)
Mk nghĩ đề có chút sai , mk làm đến đây là đc r , thông cảm nha bạn
S
18 tháng 7 2018
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{99}=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2B=1+\frac{1}{2}+...+\frac{1}{2^{98}}\)
\(2B-B=1+\frac{1}{2}+...+\frac{1}{2^{98}}-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)\)
\(B=1-\frac{1}{2^{99}}< 1\)
DM
0
NT
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
28 tháng 10 2019
\(\frac{B}{2}=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
\(\frac{B}{2}=B-\frac{B}{2}=\frac{1}{2}-\frac{1}{2^{100}}< 1\)