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Ta có:B = \(\frac{1}{2}+\frac{3}{2^2}+\frac{7}{2^3}+...+\frac{2^{100}-1}{2^{100}}=\frac{2-1}{2}+\frac{2^2-1}{2^2}+\frac{2^3-1}{2^3}+...+1-\frac{1}{2^{100}}\)
\(=1-\frac{1}{2}+1-\frac{1}{2^2}+1-\frac{1}{2^3}+...+1-\frac{1}{2^{100}}=100-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
=> \(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}\)
=> \(B=100-\left(1-\frac{1}{2^{100}}\right)=100-1+\frac{1}{2^{100}}=99+\frac{1}{2^{100}}>99\) (Đpcm)
\(A=\left(100^2+98^2+...+2^2\right)-\left(99^2+97^2+...+1^2\right)\)
\(A=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(A=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(A=100+99+98+97+...+2+1\)
Số số hạng \(100-1+1=100\) ( số hạng )
\(A=\frac{\left(100+1\right).100}{2}=5050\)
Vậy \(A=5050\)
Chúc bạn học tốt ~
\(0-\frac{2}{99}-\frac{2}{98}-...-\frac{2}{3}-1-1\)
\(=0-\left(\frac{2}{99}+\frac{2}{98}+...+\frac{2}{3}+\frac{2}{2}+\frac{2}{2}\right)\)
Đặt \(A=\frac{2}{2}+\frac{2}{2}+\frac{2}{3}+...+\frac{2}{98}+\frac{2}{99}\) , ta có:
\(A=2\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}+\frac{1}{99}\right)\).
Tự làm tiếp nha,mik có việc phải ra ngoài rồi
Ta có \(\frac{1}{2}+\frac{3}{2^2}+\frac{7}{2^3}+...+\frac{2^{100}-1}{2^{100}}\)
= \(\frac{2-1}{2}+\frac{2^2-1}{2^2}+\frac{2^3-1}{2^3}+...+\frac{2^{100}-1}{2^{100}}\)
= \(1-\frac{1}{2}+1-\frac{1}{2^2}+1-\frac{1}{2^3}+...+1-\frac{1}{2^{100}}\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)(100 hạng tử 1)
\(=100-\left(1-\frac{1}{2^{100}}\right)=100-1+\frac{1}{2^{100}}=99+\frac{1}{2^{100}}>99\)(đpcm)
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
\(\Rightarrow A< \frac{3}{4}\)
\(C=1+3+3^2+3^3+...+3^{100}\)
\(3C=3+3^2+3^3+3^4+...+3^{101}\)
\(3C-C=\left(3+3^2+3^3+3^4+...+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{100}\right)\)
\(2C=3^{101}-1\)
\(C=\frac{3^{101}-1}{2}\)
\(\frac{a_1-1}{100}=\frac{a_2-2}{99}=\frac{a_3-3}{98}=...=\frac{a_{100}-100}{1}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a_1-1+a_2-2+a_3-3+...+a_{100}-100}{1+2+3+...+100}\)\(=\)\(\frac{a_1+a_2+a_3+...+a_{100}-\left(1+2+3+...+100\right)}{1+2+3+...+100}\)
\(=\)\(\frac{10100-5050}{5050}\)vì \(1+2+3+...+100=5050\)
\(=\) \(\frac{5050}{5050}\)\(=\)\(1\)
Ta có \(\frac{a_1-1}{100}=1\Rightarrow a_1-1=100\Rightarrow a_1=101\)
\(\frac{a_2-2}{99}=1\Rightarrow a_2-2=99\Rightarrow a_2=101\)
\(\frac{a_3-3}{98}=1\Rightarrow a_3-3=98\Rightarrow a_3=101\)
\(....\)
\(\frac{a_{100}-100}{1}=1\Rightarrow a_{100}-100=1\Rightarrow a_{100}=101\)
Vậy \(a_1=a_2=a_3=....=a_{100}=101\)
A = 2100 - 299 + 298 - 297 + ... + 22 - 2
= ( 2100 + 298 + ... + 22 ) - ( 299 + 297 + ... + 2 )
= ( 2100 + 298 + ... + 22 ) - 2( 299 + 297 + ... + 2 ) + ( 299 + 297 + ... + 2 )
= 299 + 297 + ... + 2
=> 4A = 2103 + 299 + ... + 23
=> 3A = 2103 - 2
=> A = \(\frac{2^{103}-2}{3}\)
Ta có: \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(=2^{99}\left(2-1\right)+2^{97}\left(2-1\right)+...+2\left(2-1\right)\)
\(=2^{99}+2^{97}+...+2^3+2\)
\(\Leftrightarrow4A=2^{101}+2^{99}+...+2^5+2^3\)
\(\Leftrightarrow3A=2^{101}-2\)
\(\Leftrightarrow A=\dfrac{2^{101}-2}{3}\)