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NT
0
TC
1
BN
1
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
NM
3
25 tháng 2 2019
\(M=1+\frac{1}{2.\left(1+2\right)}+\frac{1}{3.\left(1+2+3\right)}+...+\frac{1}{99.\left(1+2+3+...+99\right)}\)
\(M=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{4950}\)
\(M=1-\frac{1}{4950}\)
\(M=\frac{4949}{4950}\)
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
24 tháng 2 2017
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\)
\(=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)\)
\(A=1-\frac{1}{2^{99}}\)
\(\Rightarrow B=\left(1-\frac{1}{2^{99}}\right)+\frac{1}{2^{99}}=1\)
TM
24 tháng 2 2017
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)=>\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
=>\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)\)
=>.\(A=1-\frac{1}{2^{99}}\)
=> \(B=A+\frac{1}{2^{99}}=1-\frac{1}{2^{99}}+\frac{1}{2^{99}}=1\)
20 tháng 6 2016
C = 1/3 + 1/3^2 + 1/3^3 + ... =1/3^99
=> C = 1/3^99 = 1/(3^99)
=> C < 1/2 (đpcm)
CC
20 tháng 6 2016
2A=2^101-2^100+2^98+...+2^3-2^2
3A = 2A + A
3A = 2^101 - 2 ( Cứ tính là ra , âm vs dương triệt tiêu )
A = (2^101-2) :3
B tăng tự
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
23 tháng 9 2017
\(A=\frac{1}{2}+\frac{1}{2^2}+.............+\frac{1}{2^{99}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+...........+\frac{1}{2^{98}}\)
\(\Leftrightarrow2A-A=\left(1+\frac{1}{2}+.......+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{99}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{99}}\)
\(\Leftrightarrow2^{99}.A=2^{99}-1\left(đpcm\right)\)
\(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\cdot\cdot\cdot+\left(\frac{1}{2}\right)^{99}\)
\(\Rightarrow2A=1+\frac{1}{2}+\cdot\cdot\cdot+\left(\frac{1}{2}\right)^{98}\)
\(\Rightarrow2A-A=\left[1+\cdot\cdot\cdot+\left(\frac{1}{2}\right)^{98}\right]-\left[\frac{1}{2}+\cdot\cdot\cdot+\left(\frac{1}{2}\right)^{99}\right]\)
\(\Rightarrow A=1-\left(\frac{1}{2}\right)^{99}\)
\(\Rightarrow A=1-\frac{1}{2^{99}}\)
TL:
Đặt \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+....+\left(\frac{1}{2}\right)^{99}=A\)
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A=1+\frac{1}{2}+....+\frac{1}{2^{98}}\)
\(2A-A=1-\frac{1}{2^{99}}\)
\(A=\frac{2^{99}-1}{2^{99}}\)
vậy.........