Cho D=15+\(2_{}^4+2_{}^5+2_{}^6+\ldots+2^{2020}\) . Tính giá trị của D
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\(=\left(log_{a^{-1}}a^2\right)^2+\dfrac{1}{2}.\dfrac{1}{2}log_aa\)
\(=\left(-1.2.log_aa\right)^2+\dfrac{1}{4}=4+\dfrac{1}{4}=\dfrac{17}{4}\)
Cho biết \(cosx=-\dfrac{1}{2}\)
\(sin^2x+cos^2x=1\Rightarrow sin^2x=1-cos^2x\)
\(\Rightarrow sin^2x=1-\dfrac{1}{4}=\dfrac{3}{4}\)
\(S=4sin^2x+8tan^2x\)
\(\Rightarrow S=4\left(sin^2x+2\dfrac{sin^2x}{cos^2x}\right)\)
\(\Rightarrow S=4\left(\dfrac{3}{4}+2\dfrac{\dfrac{3}{4}}{\dfrac{1}{4}}\right)\)
\(\Rightarrow S=4\left(\dfrac{3}{4}+6\right)\)
\(\Rightarrow S=4.\dfrac{27}{4}=27\)
M = 22 - ( 3 + 4 - 7 + 2020 )
= 4 - ( 7 - 7 + 2020 )
= 4 - 2020
= -2016
=> M = -2016
Đặt A=12-22+.....-20162
=> -A=22-12+42-32+62-52...+20162-20152
-A=(2-1)(2+1)+(4-3)(4+3)+(6-5)(6+5)...+(2016-2015)(2016+2015)
-A=3+7+11+...+4031
-A=[(4031-3):4+1]:2 x (3+4031)
-A=2033136
A=-2033136
Đặt A=12-22+.....-20162
=> -A=22-12+42-32+62-52...+20162-20152
-A=(2-1)(2+1)+(4-3)(4+3)+(6-5)(6+5)...+(2016-2015)(2016+2015)
-A=3+7+11+...+4031
-A=[(4031-3):4+1]:2 x (3+4031)
-A=2033136
A=-2033136
\(D=15+2^4+2^5+2^6+...+2^{2020}\\ 2D=30+2^5+2^6+2^7+...+2^{2021}\\ 2D-D=\left(30+2^5+2^6+2^7+...+2^{2021}\right)-\left(15+2^4+2^5+2^6+...+2^{2020}\right)\\ D=30+2^{2021}-15-2^4\\ D=2^{2021}+15-2^4=2^{2021}+15-16\\ D=2^{2021}-1\)
subject làm sai hả ???