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a) P = 1 + 3 + 3² + ... + 3¹⁰¹
= (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + ... + (3⁹⁹ + 3¹⁰⁰ + 3¹⁰¹)
= 13 + 3³.(1 + 3 + 3²) + ... + 3⁹⁹.(1 + 3 + 3²)
= 13 + 3³.13 + ... + 3⁹⁹.13
= 13.(1 + 3³ + ... + 3⁹⁹) ⋮ 13
Vậy P ⋮ 13
b) B = 1 + 2² + 2⁴ + ... + 2²⁰²⁰
= (1 + 2² + 2⁴) + (2⁶ + 2⁸ + 2¹⁰) + ... + (2²⁰¹⁶ + 2²⁰¹⁸ + 2²⁰²⁰)
= 21 + 2⁶.(1 + 2² + 2⁴) + ... + 2²⁰¹⁶.(1 + 2² + 2⁴)
= 21 + 2⁶.21 + ... + 2²⁰¹⁶.21
= 21.(1 + 2⁶ + ... + 2²⁰¹⁶) ⋮ 21
Vậy B ⋮ 21
c) A = 2 + 2² + 2³ + ... + 2²⁰
= (2 + 2² + 2³ + 2⁴) + (2⁵ + 2⁶ + 2⁷ + 2⁸) + ... + (2¹⁷ + 2¹⁸ + 2¹⁹ + 2²⁰)
= 30 + 2⁴.(2 + 2² + 2³ + 2⁴) + ... + 2¹⁶.(2 + 2² + 2³ + 2⁴)
= 30 + 2⁴.30 + ... + 2¹⁶.30
= 30.(1 + 2⁴ + ... + 2¹⁶)
= 5.6.(1 + 2⁴ + ... + 2¹⁶) ⋮ 5
Vậy A ⋮ 5
d) A = 1 + 4 + 4² + ... + 4⁹⁸
= (1 + 4 + 4²) + (4³ + 4⁴ + 4⁵) + ... + (4⁹⁷ + 4⁹⁸ + 4⁹⁹)
= 21 + 4³.(1 + 4 + 4²) + ... + 4⁹⁷.(1 + 4 + 4²)
= 21 + 4³.21 + ... + 4⁹⁷.21
= 21.(1 + 4³ + ... + 4⁹⁷) ⋮ 21
Vậy A ⋮ 21
e) A = 11⁹ + 11⁸ + 11⁷ + ... + 11 + 1
= (11⁹ + 11⁸ + 11⁷ + 11⁶ + 11⁵) + (11⁴ + 11³ + 11² + 11 + 1)
= 11⁵.(11⁴ + 11³ + 11² + 11 + 1) + 16105
= 11⁵.16105 + 16105
= 16105.(11⁵ + 1)
= 5.3221.(11⁵ + 1) ⋮ 5
Vậy A ⋮ 5
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a) \(A=1+2+2^2+...+2^{80}\)
\(2A=2+2^2+2^3+...+2^{81}\)
\(2A-A=2+2^2+2^3+...+2^{81}-1-2-2^2-...-2^{80}\)
\(A=2^{81}-1\)
Nên A + 1 là:
\(A+1=2^{81}-1+1=2^{81}\)
b) \(B=1+3+3^2+...+3^{99}\)
\(3B=3+3^2+3^3+...+3^{100}\)
\(3B-B=3+3^2+3^3+...+3^{100}-1-3-3^2-...-3^{99}\)
\(2B=3^{100}-1\)
Nên 2B + 1 là:
\(2B+1=3^{100}-1+1=3^{100}\)
2)
a) \(2^x\cdot\left(1+2+2^2+...+2^{2015}\right)+1=2^{2016}\)
Gọi:
\(A=1+2+2^2+...+2^{2015}\)
\(2A=2+2^2+2^3+...+2^{2016}\)
\(A=2^{2016}-1\)
Ta có:
\(2^x\cdot\left(2^{2016}-1\right)+1=2^{2016}\)
\(\Rightarrow2^x\cdot\left(2^{2016}-1\right)=2^{2016}-1\)
\(\Rightarrow2^x=\dfrac{2^{2016}-1}{2^{2016}-1}=1\)
\(\Rightarrow2^x=2^0\)
\(\Rightarrow x=0\)
b) \(8^x-1=1+2+2^2+...+2^{2015}\)
Gọi: \(B=1+2+2^2+...+2^{2015}\)
\(2B=2+2^2+2^3+...+2^{2016}\)
\(B=2^{2016}-1\)
Ta có:
\(8^x-1=2^{2016}-1\)
\(\Rightarrow\left(2^3\right)^x-1=2^{2016}-1\)
\(\Rightarrow2^{3x}-1=2^{2016}-1\)
\(\Rightarrow2^{3x}=2^{2016}\)
\(\Rightarrow3x=2016\)
\(\Rightarrow x=\dfrac{2016}{3}\)
\(\Rightarrow x=672\)
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a)Đặt \(A=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)
\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\)
\(A=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)
\(A=\dfrac{1}{3}-\dfrac{1}{12}\)
\(A=\dfrac{1}{4}\)
b)Đặt \(B=\dfrac{1}{501}+\dfrac{1}{502}+...+\dfrac{1}{1000}\)(có 500 số hạng)
\(B< \dfrac{1}{500}+\dfrac{1}{500}+...+\dfrac{1}{500}\)(có 500 số hạng)
\(B< 500\cdot\dfrac{1}{500}=1\)
\(\Rightarrow B< 1\left(đpcm\right)\)
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\(A=\frac{10}{27}+\frac{9}{16}\frac{11}{34}\)
Ta có: \(\frac{10}{27}< >\backslash\left(\frac{9}{16}< >\backslash\left(\frac{11}{34}< >Nên\backslash\left(A< >b\right)\right)\right)\backslash\left(B=\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{22}\right)\)
\(B>\frac{1}{22}+\frac{1}{22}+\frac{1}{22}+...+\frac{1}{22}=11.\frac{1}{22}=\frac{1}{2}\)
Nên \(B>\frac{1}{2}\)
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\(\dfrac{1}{12}>\dfrac{1}{22};\dfrac{1}{13}>\dfrac{1}{22};...;\dfrac{1}{21}>\dfrac{1}{22};\dfrac{1}{22}=\dfrac{1}{22}\)
\(\Rightarrow\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+...+\dfrac{1}{22}>\dfrac{1}{22}.11\) (do A có 11 số hạng)
\(\Leftrightarrow A>\dfrac{11}{22}=\dfrac{1}{2}\) ( đpcm)
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Ta có: 1/12>1/22 ; 1/13> 1/22.....1/21>1/22
Vậy: 1/12+1/13+...+1/22 > 1/22+1/22+1/22+...+1/22 = 11/22 = 1/2 (có 11 số hạng1/22).
hay: A>1/2
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Ta có:
1/2=1/22+1/22+...+1/22 có 11 p/số
A=1/12+1/13+...+1/22 có 11 p/số
Vì 1/12>1/22
1/13>1/22
.....
1/21>1/22
1/22=1/22
=>A>2
Ai thấy đúng thì !!
\(A=\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{49^2}+\frac{1}{50^2}.\)
\(\Rightarrow A< \frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{48.49}+\frac{1}{49.50}\)
\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...-\frac{1}{49}+\frac{1}{50}\)
\(A< 1+1-\frac{1}{50}=2-\frac{1}{50}< 2\)
\(\Rightarrow A< 2\left(đpcm\right)\)