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17 tháng 12 2017
\(Q=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
\(3Q=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(3Q-Q=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)
\(2Q=1-\frac{1}{3^{100}}< 1\)
\(\Rightarrow Q=\frac{1-\frac{1}{3^{100}}}{2}< \frac{1}{2}\)
9 tháng 6 2017
sửa đề câu 1 :
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
sửa đề câu 2
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
HP
5 tháng 12 2015
đặt biểu thức là A
=> \(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+..+\frac{1}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)=1-\frac{1}{3^{100}}\)
\(2A=1-\frac{1}{3^{100}}=>A=\frac{1-\frac{1}{3^{100}}}{2}\)
dễ thấy 1-1/3^100 <1
=>A<1/2(đpcm)
nhớ ****
MT
20 tháng 8 2015
\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(\Rightarrow3C=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3C-C=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^{99}}\)
\(\Rightarrow2C=1-\frac{1}{3^{99}}\)
\(\Rightarrow C=\frac{1-\frac{1}{3^{99}}}{2}
MS
4
4 tháng 2 2019
Ta có : \(VT=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
4 tháng 2 2019
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...+\frac{99}{100!}=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+\frac{5-1}{5!}+...+\frac{100-1}{100!}\)
\(=\frac{2}{1.2}-\frac{1}{2!}+\frac{3}{1.2.3}-\frac{1}{3!}+\frac{4}{1.2.3.4}-\frac{1}{4!}+\frac{5}{1.2.3.4.5}-\frac{1}{5!}+...+\frac{100}{1.2...99.100}-\frac{1}{100!}\)
\(=\frac{1}{1}-\frac{1}{2!}+\frac{1}{1.2}-\frac{1}{3!}+\frac{1}{1.2.3}-\frac{1}{4!}+\frac{1}{1.2.3.4}-\frac{1}{5!}+...+\frac{1}{1.2...99}-\frac{1}{100!}\)
\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+\frac{1}{4!}-\frac{1}{5!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
VM
6 tháng 1 2020
Đặt \(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(\Rightarrow A=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(\Rightarrow A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{100}{100!}-\frac{1}{100!}\)
\(\Rightarrow A=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(\Rightarrow A=\frac{1}{1!}-\frac{1}{100!}\)
\(\Rightarrow A=1-\frac{1}{100!}\)
Mà \(1-\frac{1}{100!}< 1.\)
\(\Rightarrow A< 1\left(đpcm\right).\)
Chúc bạn học tốt!
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
27 tháng 8 2016
\(3C=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{98}}\)
\(2C=3C-C=1-\frac{1}{3^{99}}\Rightarrow C=\left(1-\frac{1}{3^{99}}\right):2=\frac{1}{2}-\frac{1}{2.3^{99}}< \frac{1}{2}\)
Ta có: \(y=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}\Leftrightarrow3y=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{98}}\)
\(\Leftrightarrow3y-y=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{99}\right)\)
\(\Leftrightarrow2y=1-\frac{1}{3^{99}}