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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
T
2
YL
5 tháng 2 2017
Ta có: B= \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\)
=> \(\frac{1}{2}B=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{100}+\left(\frac{1}{2}\right)^{100}\)
=> B - \(\frac{1}{2}B=\left(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\right)\)
\(-\left(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{4}\right)^4+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{100}+\left(\frac{1}{2}\right)^{100}\right)\)
=> B - \(\frac{1}{2}B=\left(\frac{1}{2}+\left(\frac{1}{2}\right)^{99}\right)-\left(\left(\frac{1}{2}\right)^{100}+\left(\frac{1}{2}\right)^{100}\right)=\frac{1}{2}\)
=> B \(\times\left(1-\frac{1}{2}\right)=\frac{1}{2}\)
=> B = 1
Câu này chắc chắn đúng luôn
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
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{99}=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2B=1+\frac{1}{2}+...+\frac{1}{2^{98}}\)
\(2B-B=1+\frac{1}{2}+...+\frac{1}{2^{98}}-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)\)
\(B=1-\frac{1}{2^{99}}< 1\)
NT
2
18 tháng 2 2017
\(S=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+....+98.99}\)
\(=\frac{1+\frac{2.3}{2}+\frac{3.4}{2}+....+\frac{98.99}{2}}{1.2+2.3+3.4+....+98.99}\)
\(=\frac{\frac{1}{2}\left(1.2+2.3+3.4+....+98.99\right)}{1.2+2.3+3.4+....+98.99}\)
\(=\frac{1}{2}\)
KL
1
15 tháng 7 2017
\(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow2B=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{97}+\left(\dfrac{1}{2}\right)^{98}\)
\(\Rightarrow2B-B=\left[1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{97}+\left(\dfrac{1}{2}\right)^{98}\right]-\left[\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\right]\)
\(\Rightarrow B=1-\left(\dfrac{1}{2}\right)^{99}\)
PP
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
10 tháng 10 2015
\(\frac{N}{2}=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
\(\frac{N}{2}=N-\frac{N}{2}=\frac{1}{2}-\frac{1}{2^{100}}\Rightarrow N=1-\frac{1}{2^{99}}
BN
1
VG
5
21 tháng 2 2017
ĐỀ SAI
B=\(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+.....+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
\(2B=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+......+\left(\frac{1}{99}\right)^2+\left(\frac{1}{100}\right)^2\)
2B-B=(1/100)^2-1/2
BẢO THẰNG ĐINH ĐỨC HÙNG LÀM TIẾP Ý
HN
1
22 tháng 7 2019
A=-1++(-1)+..+-(1) có 50 số -1
=>A=-1x50=-50
B=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)
B=0+0+0+..+0
B=0
C=2^100-(2^99+2^98+...+1)
C=2^100-(2^100-1)
C=1