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\(A=2^{100}-2^{99}-...-2^2-2\)
\(2A=2^{101}-2^{100}-...-2^3-2^2\)
\(2A-A=2^{101}-2^{100}-...-2^3-2^2-2^{100}+2^{99}+...2^2+2\)
\(A=2^{101}-\left(2^{100}-2^{100}+2^{99}-2^{99}+...+2^2-2^2+-2\right)\)
\(A=2^{101}+2\)
2100 -299=21
298-297=21
=> từ 21 -> 2100 có 50 số 21
=>2100-299-298-...-21=21.50=250
\(B=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{100^2}{99.101}=\dfrac{2.3.4...100}{1.2.3...99}.\dfrac{2.3.4..100}{3.4.5...101}=100.\dfrac{2}{101}=\dfrac{200}{101}\)
Lời giải:
$S=5(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2022.2023})$
$=5(\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2023-2022}{2022.2023})$
$=5(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2022}-\frac{1}{2023})$
$=5(\frac{1}{2}-\frac{1}{2023})=\frac{10105}{4046}$
\(S=\dfrac{2^2}{1.2}+\dfrac{2^2}{2.3}+\dfrac{2^2}{3.4}+...+\dfrac{2^2}{2022.2023}\)
\(S=2^2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)
\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2023}\right)\)
\(S=2^2.\dfrac{2022}{2023}\)
\(S=\dfrac{2^2.2022}{2023}=\dfrac{8088}{2023}\)