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Bài làm:
Ta có: \(\left(a-b\right)\left(a+b\right)=a^2+ab-ab-b^2=a^2-b^2\)
Áp dụng vào tính: (đề đoạn cuối thiếu bình phương của 99 nhé)
Ta có: \(\left(1-\frac{1}{2^2}\right)\cdot\left(1-\frac{1}{3^2}\right)\cdot...\cdot\left(1-\frac{1}{99^2}\right)\)
\(=\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot...\cdot\frac{99^2-1}{99^2}\)
\(=\frac{\left(2-1\right)\left(2+1\right)}{2^2}\cdot\frac{\left(3-1\right)\left(3+1\right)}{3^2}\cdot...\cdot\frac{\left(99-1\right)\left(99+1\right)}{99^2}\)
\(=\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}\cdot...\cdot\frac{98.100}{99^2}\)
\(=\frac{\left(1\cdot2\cdot...\cdot98\right)\cdot\left(3\cdot4\cdot...\cdot100\right)}{\left(2\cdot3\cdot...\cdot99\right)\cdot\left(2\cdot3\cdot...\cdot99\right)}=\frac{100}{2\cdot99}=\frac{50}{99}\)
Ta có:
A=1/3 - 2/3^2+3/3^3 - 4/3^4+ ... - 100/3^100
=>3A=1 -2/3 +3/3^2 - 4/3^3+ ... - 100/3^99
=>4A=A+3A=1-1/3+1/3^2-1/3^3+...-1/3^99 - 100/3^100
=>12A=3.4A=3-1+1/3-1/3^2+...-1/3^98 - 100/3^99
=>16A=12A+4A=3-1/3^99-100/3^99-100/3^1...
<=>16A=3-101/3^99-100/3^100
<=>A=3/16-(101/3^99+100/3^100)/16 < 3/16
Suy ra A<3/16
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\rightarrow A=\frac{3}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\rightarrow A=\frac{7}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\rightarrow A=\frac{15}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\rightarrow A=\frac{31}{32}+\frac{1}{64}\)
\(\rightarrow A=\frac{63}{64}\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\Rightarrow64A=32+16+8+4+2+1\Rightarrow64A=63\Rightarrow A=\frac{63}{64}\)
A = 1.2 + 2.3 + 3.4 + 4.5 + ......................... + 99.100
3A = 1.2.3 + 2.3.3 + 3.4.3 +.................. + 99.100.3
3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ............... + 99.100.(101 - 98)
3A =1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ................. + 99.100.101 - 98.99.100
3A = 99.100.101
A = 99.100.101 : 3
A = 33.100.101
A = 3300.101
A = 30300
a) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.......+\frac{1}{2^{99}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{98}}\)
\(\Rightarrow2A-A=A=1-\frac{1}{2^{99}}\)
b) \(B=\frac{1}{2^2}+\frac{1}{2^4}+...........+\frac{1}{2^{100}}\)
\(\Rightarrow2^2.B=4B=1+\frac{1}{2^2}+......+\frac{1}{2^{98}}\)
\(\Rightarrow4B-B=3B=1-\frac{1}{2^{100}}\)
\(\Rightarrow b=\frac{1-\frac{1}{2^{100}}}{3}\)
a) Ta có A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
=> 2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)\)
=> A = \(1-\frac{1}{2^{99}}\)
b) Ta có B = \(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}\)
=> 22B = \(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}=4B\)
=> 4B - B = \(\left(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}\right)\)
=> 3B = \(1-\frac{1}{2^{100}}\)
=> B = \(\frac{1}{3}-\frac{1}{2^{100}.3}\)