C= (- 2/5) ^ 0 + (- 2/5) ^ 2 + (- 2/5) ^ 4 +...+(- 2/5 ) ^ 100
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c: \(=\dfrac{3}{2}\cdot1-1-20=\dfrac{3}{2}-21=\dfrac{-39}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c: \(=\dfrac{3}{2}-1-21=\dfrac{3}{2}-21=\dfrac{-39}{2}\)
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\(a,A=2^0+2^1+2^2+....+\)\(2^{2010}\)
\(\Rightarrow2A=2^1+2^2+2^3+....+2^{2011}\)
\(2A-A=\left(2^1+2^2+2^3+...+2^{2011}\right)-\left(2^0+2^1+2^2+...+2^{2010}\right)\)
\(A=2^{2011}-2^0\)
\(A=2^{2011}-1\)
\(b,B=1+3+3^2+...+3^{100}\)
\(\Rightarrow3B=3+3^2+3^3+...+3^{101}\)
\(3B-B=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)
\(2B=3^{101}-1\)
\(\Rightarrow B=\frac{3^{101}-1}{2}\)
\(c,C=4+4^2+4^3+...+4^n\)
\(\Rightarrow4C=4^2+4^3+4^4+...+4^{n+1}\)
\(4C-C=\left(4^2+4^3+4^4+...+4^{n+1}\right)-\left(4+4^2+4^3+...+4^n\right)\)
\(3C=4^{n+1}-4\)
\(\Rightarrow C=\frac{4^{n+1}-4}{3}\)
\(d,D=1+5+5^2+...+5^{2000}\)
\(\Rightarrow5D=5+5^2+5^3+...+5^{2001}\)
\(5D-D=\left(5+5^2+5^3+...+5^{2001}\right)-\left(1+5+5^2+...+5^{2000}\right)\)
\(4D=5^{2001}-1\)
\(\Rightarrow D=\frac{5^{2001}-1}{4}\)
b)
B=1+3+3^2+3^3+..+3^100
=> 3B = 3 + 3^2 + 3^3 + ...+ 3^101
=> 3B - B = ( 3 + 3^2 + 3^3 + ...+ 3^101) - (1+3+3^2+3^3+..+3^100)
=> 2B = 3^101 - 1
=> B =( 3^101 - 1) / 2
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a) 3. 5 ^ 2 + 15. 2 ^2 - 26 : 2
= 15 ^ 2 + 15. 4 - 13
= 225 + 60 - 13
= 272
b) 5 ^ 3. 2 - 100 : 4 + 2 ^ 3. 5
= 125. 2 - 25 + 40
= 250 - 25 + 40
= 265
c) 6 ^ 2 : 9 + 50. 2 - 3 ^ 3. 3
= 36 : 9 + 100 - 81
= 4 + 100 - 81
= 23
d) 3 ^ 2. 5 + 2 ^ 3 . 10 - 81 : 3
= 45 + 8 . 10 - 27
= 53 + 80 - 27
= 160
e) 5 ^ 13 : 5 ^ 10 - 25. 2 ^ 2
= 1220703125 : 9765625 - 25. 4
= 125 - 100
= 25
f) 20 : 2 ^ 2 + 5 ^ 9 : 5 ^ 8
= 20 : 4 + 1953125 : 390625
= 5 + 5
= 10
g) 100 : 5 ^ 2 + 7. 3 ^ 2
= 100 : 25 + 7 . 9
= 4 + 63
= 67
h) 84 :4 + 3 ^ 9 : 3 ^ 7 + 5 ^ 0
= 21 + 19683 : 2187 + 1
= 21 + 9 + 1
= 31
i) 29 - [ 16 + 3. ( 51 - 49) ]
= 29 - [ 16 + 3. 2 ]
= 29 - 22
= 7
Mình làm xong rồi nhé!
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