S=2+2mu2+2mu3+....+2mu100;hay chung to rang S chia het cho 15 va chu so tan cung cua S
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\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{20}}\)
=> \(2S=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{19}}\)
=> \(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\right)\)
=> \(S=1-\frac{1}{2^{20}}\)
\(A=2^{100}-2^{99}-...-2^2-2\)
\(2A=2^{101}-2^{100}-...-2^3-2^2\)
\(2A-A=2^{101}-2^{100}-...-2^3-2^2-2^{100}+2^{99}+...2^2+2\)
\(A=2^{101}-\left(2^{100}-2^{100}+2^{99}-2^{99}+...+2^2-2^2+-2\right)\)
\(A=2^{101}+2\)
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2100 -299=21
298-297=21
=> từ 21 -> 2100 có 50 số 21
=>2100-299-298-...-21=21.50=250