P
3
K
Khách
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KV
1
27 tháng 9 2018
Ta có:
\(3D=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(3D-D=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)
\(2D=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
Đặt \(E=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3E=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3E-E=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(2E=3-\frac{1}{3^{99}}< 3\)
\(E< \frac{3}{2}\)
\(2D< \frac{3}{2}-\frac{1}{3^{100}}< \frac{3}{2}\)
\(D< \frac{3}{4}\)
Vậy...
5 tháng 11 2016
\(C=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)
\(3C=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(3C-C=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)
\(2C=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6C=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6C-2C=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4C=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4C=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4C=3-\frac{203}{3^{100}}< 3\)
\(\Rightarrow C< \frac{3}{4}\left(đpcm\right)\)
NT
1
D
28 tháng 10 2018
Ta có:
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)
\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6A=3+1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow4A=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}=3-\frac{203}{3^{100}}\)
\(\Rightarrow A=\frac{3-\frac{203}{3^{100}}}{4}=\frac{3}{4}-\frac{203}{3^{100}.4}< \frac{3}{4}\Rightarrowđpcm\)
Vậy \(A< \frac{3}{4}\)
M
2
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
25 tháng 11 2019
\(\frac{A}{3}=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{100}{3^{101}}.\)
\(\frac{2A}{3}=A-\frac{A}{3}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{5^5}+...+\frac{1}{3^{100}}-\frac{100}{3^{101}}\)
Đặt \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(\frac{B}{3}=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{101}}\)
\(\frac{2B}{3}=B-\frac{B}{3}=\frac{1}{3}-\frac{1}{3^{101}}\Rightarrow B=\frac{3^{100}-1}{2}\)
\(\frac{2A}{3}=B-\frac{100}{3^{101}}=\frac{3^{100}-1}{2}-\frac{100}{3^{101}}=\frac{3^{201}-3^{101}-200}{2.3^{101}}\)
\(\Rightarrow A=\frac{3^{201}-3^{101}-200}{4.3^{100}}\)
KH
0
BT
0
KM
0
DB
3
11 tháng 5 2015
Đặt A \(=\) \(\frac{1}{3}+\frac{2}{3^2}+...+\frac{100}{3^{100}}\)
=> 3A\(=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
=> 3A- A \(=\) 2A \(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
Đặt B \(=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)=>\(3B=3+1+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)
=> 2B \(=3-\frac{1}{3^{99}}<3\) =>B < \(\frac{3}{2}\) => 2A< \(\frac{3}{2}\) => A < \(\frac{3}{4}\)
ĐÚNG CÁI NHÉ BẠN
năm nay mới lên lớp 7
đặt B = 100/3 + 100/3^2 + 100/3^3 + 100/3^4
B = 100 x ( 1/3 + 1/3^2 +1/3^3 +1/3^4 )
đặt A = 1/3 + 1/3^2 + 1/3^3 + 1/3^4
3A = 1 + 1/3 + 1/3^2 + 1/3^3
3A - A = 1 - 1/3^4
2A = 81/81 - 1/81 = 80/81
A = 80/81 : 2 = 40/81
thay A vào B ta có B= 100 x 40/81 = 4000/81
vậy B = 4000 /81
chúc bạn hok tốt