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câu g)
\(G=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right).\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{120}{121}\)
\(=\frac{3.\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{2.2.3.3.4.4.5.5....11.11}\)
\(=\frac{12}{3}=4\)
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 :
\(D=2^{100}-2^{99}+2^{98}-.....-2\) \(\Rightarrow2D=2^{101}-2^{100}+2^{99}-.....-2^2\)
\(\Rightarrow2D+D=2^{101}-2=3D\) \(\Rightarrow D=\frac{2^{101}-2}{3}\)
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(2A=1+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)-\frac{100}{3^{100}}\)
Ta thấy biểu thức trong dấu ngoặc nhỏ hơn 1/2 ( tự chứng minh ) nên 2A < 1 + 1/2
\(\Rightarrow A< \frac{3}{4}\)
\(C=1+3+3^2+3^3+...+3^{100}\)
\(3C=3+3^2+3^3+3^4+...+3^{101}\)
\(3C-C=\left(3+3^2+3^3+3^4+...+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{100}\right)\)
\(2C=3^{101}-1\)
\(C=\frac{3^{101}-1}{2}\)
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 \(34x+17y⋮17\)
\(\Rightarrow2x+3y⋮17\)