A 2+22+....+219+220
B 2+22+....+299+2100
C 3+32+.....+39+310
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a: \(A=4+2^2+2^3+...+2^{20}\)
=>\(2A=8+2^3+2^4+...+2^{21}\)
=>\(2A-A=2^{21}+2^{20}+...+2^4+2^3+8-2^{20}-2^{19}-...-2^3-2^2-4\)
\(=2^{21}+8-2^2-4=2^{21}\)
=>\(A=2^{21}\) là lũy thừa của 2
b:
\(B=3+3^2+3^3+...+3^{100}\)
=>\(3B=3^2+3^3+...+3^{101}\)
=>\(2B=3^{101}-3\)
=>\(2B+3=3^{101}\) là lũy thừa của 3
a) \(A=1+2+2^2+...+2^{50}\)
\(\Rightarrow2A=2+2^2+...+2^{51}\)
\(\Rightarrow A=2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}=2^{51}-1\)
b) \(B=1+3+3^2+...+3^{100}\)
\(\Rightarrow3B=3+3^2+...+3^{101}\)
\(\Rightarrow2B=3B-B=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}=3^{101}-1\)
\(\Rightarrow B=\dfrac{3^{101}-1}{2}\)
c) \(C=5+5^2+...+5^{30}\)
\(\Rightarrow5C=5^2+5^3+...+5^{31}\)
\(\Rightarrow4C=5C-C=5^2+5^3+...+5^{31}-5-5^2-...-5^{30}=5^{31}-5\)
\(\Rightarrow C=\dfrac{5^{31}-5}{4}\)
d) \(D=2^{100}-2^{99}+2^{98}-...+2^2-2\)
\(\Rightarrow2D=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)
\(\Rightarrow3D=2D+D=2^{101}-2^{100}+2^{99}-...+2^3-2^2+2^{100}-2^{99}+...+2^2-2=2^{101}-2\)
\(\Rightarrow D=\dfrac{2^{101}-2}{3}\)
\(A=2+2^2+...+2^{20}\)
\(2A=2^2+2^3+...+2^{21}\)
\(2A-A=2^2+2^3+...+2^{21}-2-2^2-...-2^{20}\)
\(A=2^{21}-2\)
___________
\(B=5+5^2+...+5^{50}\)
\(5B=5^2+5^3+...+5^{51}\)
\(5B-B=5^2+5^3+...+5^{51}-5-5^2-...-5^{50}\)
\(4B=5^{51}-5\)
\(B=\dfrac{5^{51}-5}{4}\)
___________
\(C=1+3+3^2+...+3^{100}\)
\(3C=3+3^2+...+3^{101}\)
\(3C-C=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}\)
\(2C=3^{101}-1\)
\(C=\dfrac{3^{101}-1}{2}\)
a: \(2^3-5^3:5^2+12\cdot2^2\)
\(=8-5+48\)
\(=51\)
b: \(5\cdot\left[\left(85-35:7\right):8+90\right]-5\)
\(=5\cdot\left[10+90\right]-5\)
=495
a: Tổng các số hạng là:
\(\dfrac{\left(220+1\right)\cdot220}{2}=24310\)
Ta có: A+1=2x
\(\Leftrightarrow2x=24311\)
hay \(x=\dfrac{24311}{2}\)
A = 2 + 22 + 23 + 24 + ... + 219 + 220
A = (2 + 22) + (23 + 24) +... + (219 + 220)
A = 2.(1+2) + 23.(1 + 2) +... + 219.(l + 2)
A = 2.3 + 23.3 +...+ 219.3 Do đó A chia hết cho 3
a: 2A=2^2+2^3+...+2^21
=>A=2^21-2
b: B=2+2^2+...+2^100
=>2B=2^2+2^3+...+2^101
=>B=2^101-2
c: C=3+3^2+...+3^10
=>3C=3^2+3^3+...+3^11
=>2C=3^11-3
=>C=(3^11-3)/2
`A = 2 + 2^2 + ... + 2^20`
`=> 2A = 2^2 + 2^3 + ... +2^21`
`=> 2A-A = (2^2 + 2^3 + ... + 2^21) - (2 + 2^2 + ... +2^20)`
`=> A = 2^21 - 2`
`B = 2 + 2^2 + ... + 2^99 + 2^100`
`=>2B = 2^2 + 2^3 + ... + 2^100 + 2^101`
`=> 2B-B = (2^2 + 2^3 + ... + 2^101)- (2 + 2^2 + ... + 2^100)`
`=> B = 2^101 - 2`
`C = 3 + 3^2 + .... + 3^10`
`=>3C = 3^2 + 3^3 + ... +3^11`
`=>3C - C = (3^2 + 3^3 + ... +3^11) - (3 + 3^2 + .... + 3^10)`
`=> 2C = 3^11 - 3`
`=> C = (3^11 - 3)/2