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Ta có:
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)
\(\Rightarrow2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
Lấy \(2A-A\), ta có:
\(2A-A=A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{2^9}-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)
\(=\left(1-\frac{1}{2^{10}}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{2^9}-\frac{1}{2^9}\right)\)
\(=1-\frac{1}{2^{10}}\)
\(=1-\frac{1}{1024}\)
\(=\frac{1023}{1024}\)
Vậy \(A=\frac{1023}{1024}\)
a: =-45*24+120
=-1080+120
=-960
b: =134(-1+51-48)
=134*2
=268
c: =-41*59-41*2+59*41-59*2
=-2(41+59)
=-200
1)
a) -(2+5) = -2 - 5 = -7
b) +(-3+6) = -3 + 6 = 3
c) (-50+3) = -50 + 3 = -47
d) -(-2+3) = 2 - 3 = -1
e) -(10-3) = -10 + 3 = -7
f) -(-3)-(-3+1) = 3 + 3 - 1 = 5
g) (-5)+(-2+10) = -5 - 2 + 10 = 3
2)
a) -50+120+(-150)-20+30
= -(50 + 20) + (120 + 30 - 150)
= -70
b) 265-70+(-65)-30+15
= (265 - 65) - (70 + 30) + 15
= 200 - 100 + 15 = 115
c) -17+185-183+(-85)-63
= (185 - 85) - (183 + 17) - 63
= 100 - 200 - 63 = -163
d) -30+60+(-170)-260+19
= -(170 + 30) - (260 - 60) + 19
= -200 - 200 + 19 = -381
\(\Rightarrow4C=4^2+4^3+...+4^{n+1}\)
\(\Rightarrow4C-C=\left(4^2+4^3+...+4^{n+1}\right)-\left(4+4^2+...+4^n\right)\)
\(\Rightarrow3C=4^{n+1}-4\)
\(\Rightarrow C=\frac{4^{n+1}-4}{3}\)
\(4C=4^2+4^3+4^4+...+4^{n+1}\)
\(4C-C=4^2+4^3+...+4^{n+1}-4-4^2-...-4^n\)
\(3C=4^{n+1}-4\)
\(C=\frac{4^{n+1}-4}{3}\)