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C=1+3+32+.............+3100
C=\(\frac{3C-C}{2}\)
3C=3+32+33+.............+399+3100+3101
C=1+3+32+..................+399+3100
3C-C=(3+32+33+.............+399+3100+3101)-(1+3+32+..................+399+3100)
Triệt tiêu các số hạng co giá trị tuyệt đối bằng nhau, ta được:
2C=-1+3100
\(\Rightarrow C=\frac{3^{100}-1}{2}\)
D=\(\frac{2D+D}{3}\)
2D=2101-2100+299-298+..............+23-22
D=2100-299+298-297+............+22-2
2D+D=2101-2100+299-298+..............+23-22+2100-299+298-297+............+22-2
Triệt tiêu các số hạng có giá trị tuyệt đối bằng nhau, ta được:
3D=2101-2
\(\Rightarrow D=\frac{2^{101}-2}{3}\)
B=\(\frac{3}{1\times4}+\frac{5}{4\times9}+\frac{7}{9\times16}+.........+\frac{19}{81\times100}\)
Quan sát biểu thức, ta có nhận xét:
4-1=3;
9-4=5;
16-9=7;
.......;100-81=19
=> Hiệu hai số ở mẫu bằng giá trị ở tử
\(\Rightarrow B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+.......+\frac{1}{81}-\frac{1}{100}\)
\(\Rightarrow B=1-\frac{1}{100}\)
\(B=\frac{99}{100}< \frac{100}{100}\)
Vậy B<1
Ta có: \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(\frac{1}{3}A=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{100}{3^{101}}\)
\(\frac{2}{3}A=\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)-\left(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{100}{3^{101}}\right)\)
\(\frac{2}{3}A=\frac{1}{3}+\left(\frac{2}{3^2}-\frac{1}{3^2}\right)+\left(\frac{3}{3^3}-\frac{2}{3^3}\right)+...+\left(\frac{100}{3^{100}}-\frac{99}{3^{100}}\right)-\frac{100}{3^{101}}\)
\(=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{100}{3^{101}}\)
Đặt: \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(\Rightarrow\frac{1}{3}B=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{101}}\)
\(\Rightarrow\frac{2}{3}B=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)-\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{101}}\right)\)
\(=\frac{1}{3}-\frac{1}{3^{101}}\)\(\Leftrightarrow B=\left(\frac{1}{3}-\frac{1}{3^{101}}\right):\frac{2}{3}=\left(\frac{1}{3}-\frac{1}{3^{101}}\right).\frac{3}{2}\)
Thay \(B\) vào \(\frac{2}{3}A\), ta có: \(\frac{2}{3}A=\frac{3}{2}\left(\frac{1}{3}-\frac{1}{3^{101}}\right)-\frac{100}{3^{101}}\)
\(\Rightarrow A=\left[\frac{3}{2}\left(\frac{1}{3}-\frac{1}{3^{101}}\right)-\frac{100}{3^{101}}\right]:\frac{2}{3}=\frac{9}{4}\left(\frac{1}{3}-\frac{1}{3^{101}}\right)-\frac{150}{3^{101}}\)
\(A=\frac{3}{4}-\frac{9}{4}.\frac{1}{3^{101}}-\frac{150}{3^{101}}\Rightarrow A< \frac{3}{4}\)
Vậy \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}< \frac{3}{4}\)(ĐPCM)
Xong.
1) \(+2x+3y⋮17\)
\(\Rightarrow26x+39y⋮17\)
\(\Rightarrow\left(9x+5y\right)+17x+34y⋮17\)
Mà \(17x+34y⋮17\)
\(\Rightarrow9x+5y⋮17\)
\(+9x+5y⋮17\)
\(\Rightarrow36x+20y⋮17\)
\(\Rightarrow\left(2x+3y\right)+34x+17y⋮17\)
Mà \(34x+17y⋮17\)
\(\Rightarrow2x+3y⋮17\)
a, \(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(2C=1-\frac{1}{3^{99}}\)
\(C=\frac{1}{2}-\frac{1}{2.3^{99}}< \frac{1}{2}\)(đpcm)
b, Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6A-2A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{397}{3^{100}}\)
\(A=\frac{3}{4}-\frac{397}{4.3^{100}}< \frac{3}{4}\)(đpcm)
\(E<\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}=\left(\frac{1}{3}-\frac{1}{3^{101}}\right):2<\frac{3}{4}\)
bài nhà nữa thôi nha
đặt:
\(M=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{4^4}+..+\frac{99}{3^{99}}+\frac{100}{3^{100}}\)
do đó:
\(3M=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{99}{3^{98}}+\frac{100}{3^{99}}\)
=>3M-M=2M=\(1+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)-\frac{100}{3^{100}}\)
ta thấy bthuc trong ngoặc nhỏ hơn 1/2
=>2M<1+1/2
hay M<3/4