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Ta có: 2E= 2+2^2+2^3+2^4+...+2^10
2E - E = (2+2^2+2^3+2^4+...+2^10) - (1+2+2^2+2^3+...+2^9)
E = 2^10-1
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2016}}\)\(\Leftrightarrow2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{2016}}\right)\)\(\Leftrightarrow2A-A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{2016}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+.............+\frac{1}{2^{2016}}\right)\)
\(\Leftrightarrow A=2-\frac{1}{2^{2016}}\)
(2^2 + 2) + (2^3 + 2^4) +...........+(2^11 + 1)
= 2. (2+1) + 2^3. (2+1) + ........ + 2^9.(2+1) +(2^11+1)
= 2. 3 + 2^3. 3 + ..... + 2^9. 3 + (2^11 +1)
Vì 3 chia hết cho 3
=> A chia hết cho 3
\(2A=2\left(1+\frac{1}{2}+...+\frac{1}{2^{2016}}\right)\)
\(2A=2+1+...+\frac{1}{2^{2015}}\)
\(2A-A=\left(2+1+...+\frac{1}{2^{2015}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2016}}\right)\)
\(A=2-\frac{1}{2^{2016}}\)
S = 1 + 2 + 22 + 23 +...+ 29
2S = 2 + 22 + 23+...+ 29 + 210
2S - S = 210 - 1
S = 210 - 1
P = 5.20 = 5 < 7 = 23 - 1 < 210 -1 = S
S > P
đặt A=1+2+2^2+2^3+2^4+2^5+2^6+2^7
2A=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8
2A-A=(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8)-(1+2+2^2+1^3+2^4+2^5+2^6+2^7)
A=2^8-1
A=256-1=255
255 chia hết cho 3
nên 1+2+2^2+2^3+2^4+2^5+2^6+2^7 cũng chia hết cho 3
\(A=1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}\)
\(2A=2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}\)
\(\Rightarrow A=2A-A=2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}-\left(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}\right)\)
\(=2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}-1-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2020}}\)
\(=1+\frac{1}{2}-\frac{1}{2^{2020}}=\frac{2^{2020}+2^{2019}-1}{2^{2020}}\)