Tính S:
S = 3 + 3/2 + 3/2^2 + 3/2^3 + ... + 3/2^9
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\(S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)
\(S=1+\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\right)\)
đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}\)
thay A=\(1-\frac{1}{2^{100}}\)vào S ta có: \(S=1+1-\frac{1}{2^{100}}=2-\frac{1}{2^{100}}\)
a: A=3^2(1^2+2^2+...+10^2)
=9*385
=3465
b: B=2^3(1^3+2^3+...+10^3)
=8*3025
=24200
`9/2xx7/3-4/3xx9/2`
`=9/2xx(7/3-4/3)`
`=9/2xx3/3`
`=9/2xx1`
`=9/2`
Ta có: S = 3+3/2+3/2^2+3/2^3+...+3/2^9
1/2.S = 3/2+3/2^2+3/2^3+3/2^4+...+3/2^10
\(\Rightarrow\) S-1/2.S = 3 - 3/2^10
\(\Rightarrow\) 1/2.S = 3 - 3/2^10
\(\Rightarrow\) S = (3 - 3/2^10) : 1/2
\(\Rightarrow\) S = 6 - 6/2^10
Nếu đúng thì cho mk biết nha
TH1: \(A=\left(9-\frac35+\frac23\right)-\left(7+\frac75-\frac32\right)-\left(3-\frac95+\frac52\right)\)
\(=\left(\frac{135}{15}-\frac{9}{15}+\frac{10}{15}\right)-\left(\frac{70}{10}+\frac{14}{10}-\frac{15}{10}\right)-\left(\frac{30}{10}-\frac{18}{10}+\frac{25}{10}\right)\)
\(=\frac{136}{15}-\frac{69}{10}-\frac{37}{10}=\frac{136}{15}-\frac{106}{10}=\frac{272}{30}-\frac{318}{30}=-\frac{46}{30}=-\frac{23}{15}\)
TH2: \(A=\left(9-\frac35+\frac23\right)-\left(7+\frac75-\frac32\right)-\left(3-\frac95+\frac52\right)\)
\(=9-\frac35+\frac23-7-\frac75+\frac32-3+\frac95-\frac52\)
\(=\left(9-7-3\right)+\left(-\frac35-\frac75+\frac95\right)+\left(\frac32-\frac52\right)+\frac23=-1-\frac15-1+\frac23\)
\(=-2-\frac15+\frac23=-\frac{30}{15}-\frac{3}{15}+\frac{10}{15}=-\frac{23}{15}\)