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\(S_1=1+2+2^2+2^3+..+2^{63}\\ \Rightarrow2S_1=2+2^2+2^3+2^4+...+2^{64}\\ \Rightarrow S_1-2S_1=1-2^{64}\\ \Rightarrow-S_1=1-2^{64}\\ \Rightarrow S_1=2^{64}-1.\)
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a) \(A=2+2^2+2^3+...+2^{2017}\)
\(2A=2^2+2^3+2^4+...+2^{2018}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)\)
\(A=2^{2018}-2\)
b) \(C=1+3^2+3^4+...+3^{2018}\)
\(3^2\cdot C=3^2+3^4+3^6+...+3^{2020}\)
\(9C-C=\left(3^2+3^4+3^6+...+3^{2020}\right)-\left(1+3^2+3^4+...+3^{2018}\right)\)
\(8C=3^{2020}-1\)
\(\Rightarrow C=\dfrac{3^{2020}-1}{8}\)
\(Toru\)
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B = 2 3 + 3. 1 9 0 − 2 − 2 .4 + − 2 2 : 1 2 .8 = 8 + 3.1 − 1 4 .4 + 4 : 1 2 .8 = 10 + 64 = 74
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2 3 + 3. 1 2 0 − 1 + − 2 2 : 1 2 − 8 = 8 + 3 − 1 + 4 : 1 2 − 8 = 2 + 8 = 10
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\(\frac{4^{50}.3^2}{2^{23}.2^{22}}\)
\(=\frac{\left(2^2\right)^{50}.3^2}{2^{23+22}}\)
\(=\frac{2^{100}.3^2}{2^{45}}\)
\(=2^{55}.3^2\)
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A=\(2^2-9^3+4^{-2}.16-2.5^2\)
\(=4-729+1-50=-774\)
B=\(\left(2^3.2\right).\dfrac{1}{2}+3^{-2}.3^2-7.1+5\)
\(B=2^4.\dfrac{1}{2}+1-7+5=8+1-7+5=7\)
`#3107.101107`
Đặt $A = 1 + 2 + 2^2 + 2^3 + ... + 2^{50}$
$2A = 2 + 2^2 + 2^3 + ... + 2^{51}$
$2A - A = (2 + 2^2 + 2^3 + ... + 2^{51}) - (1 + 2 + 2^2 + ... + 2^{50})$
$A = 2 + 2^2 + 2^3 + ... + 2^{51] - 1 - 2 - 2^2 - ... - 2^{50}$
$A = 2^{51} - 1$
Vậy, `A =` $2^{51} - 1.$