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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 \(34x+17y⋮17\)
\(\Rightarrow2x+3y⋮17\)
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
câu g)
\(G=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right).\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{120}{121}\)
\(=\frac{3.\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{2.2.3.3.4.4.5.5....11.11}\)
\(=\frac{12}{3}=4\)
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{100}\left(1+2+...+100\right)\)
\(=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+...+\frac{1}{100}\cdot\frac{100.101}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{101}{2}\)
\(=\frac{1}{2}\left(2+3+...+101\right)=\frac{1}{2}\cdot\frac{100.103}{2}=25.103=2575\)
\(3+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+3+...+100}\)
\(=3+3.\left(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\right)\)
\(=3+3.\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{5050}\right)\)
\(=3+3.\frac{1}{2}.\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\right)\)
\(=3+\frac{3}{2}.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)\)
\(=3+\frac{3}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=3+\frac{3}{2}.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=3+\frac{3}{2}.\frac{99}{202}\)
\(=3+\frac{297}{404}\)
\(=\frac{1509}{404}\)
chỗ 3+3/2(1/6+..)
bn nhìn nhầm rồi
đáng lẽ: 3+(1/6+,.....) chứ nâk
Ta có :
\(A=\frac{1}{3}+\frac{2}{3^2}+......+\frac{100}{3^{100}}\) \(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+.....+\frac{100}{3^{99}}\)
\(\Rightarrow3A-A=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)= 2A
Đặt \(B=1+\frac{1}{3}+...+\frac{1}{3^{99}}\) \(\Rightarrow3B=3+1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{98}}\)
\(\Rightarrow3B-B=3-\frac{1}{3^{99}}=2B\) \(\Rightarrow B=\frac{3}{2}-\frac{1}{3^{99}.2}\)
\(\Rightarrow2A=\frac{3}{2}-\frac{1}{3^{99}.2}-\frac{100}{3^{100}}\)\(\Rightarrow A=\frac{3}{4}-\frac{1}{3^{99}.4}-\frac{100}{3^{100}}< \frac{3}{4}\Rightarrow\left(đpcm\right)\)
Ta có :
\(C=1+3+3^2+....+3^{100}\) \(\Rightarrow C-1=3+3^2+....+3^{100}\)
\(\Rightarrow3\left(C-1\right)=3^2+3^3+.....+3^{101}\)\(\Rightarrow3C-3-\left(C-1\right)=3^{101}-3\)
\(\Rightarrow2C-2=3^{101}-3\Rightarrow2C=3^{101}-1\)\(\Rightarrow C=\frac{3^{101}-1}{2}\)
Ta có :
\(D=2^{100}-2^{99}+2^{98}-.....-2\) \(\Rightarrow2D=2^{101}-2^{100}+2^{99}-.....-2^2\)
\(\Rightarrow2D+D=2^{101}-2=3D\) \(\Rightarrow D=\frac{2^{101}-2}{3}\)
a)S=2+22+23+...+2100
2S=2(2+22+23+...+2100)
2S=22+23+...+2101
2S-S=(22+23+...+2101)-(2+22+23+...+2100)
S=2101-2
b)\(P=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(3P=3\left(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{100}}\right)\)
\(3P=1+\frac{1}{3}+...+\frac{1}{3^{99}}\)
\(3P-P=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)\)
\(2P=1-\frac{1}{3^{100}}\)
\(P=\left(1-\frac{1}{3^{100}}\right):2\)
cái này lớp 6 mà bạn :D