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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
\(1,Y=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\\ Y=\left(1+3+3^2\right)\left(1+3^3+...+3^{96}\right)\\ Y=13\left(1+3^3+...+3^{96}\right)⋮13\\ 2,A=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{2018}+3^{2019}\right)\\ A=\left(1+3\right)\left(1+3^2+...+3^{2019}\right)\\ A=4\left(1+3^2+...+3^{2019}\right)⋮4\\ 3,\Leftrightarrow2\left(x+4\right)=60\Leftrightarrow x+4=30\Leftrightarrow x=36\)
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:
\(8^x-1=2^{2016}-1\)
\(\Rightarrow\left(2^3\right)^x-1=2^{2016}-1\)
\(\Rightarrow2^{3x}-1=2^{2016}-1\)
\(\Rightarrow2^{3x}=2^{2016}\)
\(\Rightarrow3x=2016\)
\(\Rightarrow x=\dfrac{2016}{3}\)
\(\Rightarrow x=672\)
Ta có 1/2.2<1/1.2
1/3.3<1/2.3
1/4.4<1/3.4
.........................
1/20.20<1/19.20
=>1/2.2+1/3.3+1/4.4+...+1/20.20<1/1.2+1/2.3+1/3.4+...+1/19.20
=>A<1/1-1/2+1/2-1/3+1/3-1/4+...+1/19-1/20
=>A<1/1-1/20
=>A<20/20-1/20
=>A<19/20<20/20=1
=>A<1
Vậy A<1
\(A=\left(1+3\right)+3^2\left(1+3\right)+...+3^{2018}\left(1+3\right)\)
\(=4\left(1+3^2+...+3^{2018}\right)⋮4\)
a)\(\dfrac{1}{2^2}<\dfrac{1}{1.2}\)
\(\dfrac{1}{3^3}<\dfrac{1}{2.3}\)
\(...\)
\(\dfrac{1}{8^2}<\dfrac{1}{7.8}\)
Vậy ta có biểu thức:
\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)
\(B= 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(B<1-\dfrac{1}{8}=\dfrac{7}{8}<1\)
Vậy B < 1 (đpcm)
Giải:
a) Ta có:
1/22=1/2.2 < 1/1.2
1/32=1/3.3 < 1/2.3
1/42=1/4.4 < 1/3.4
1/52=1/5.5 < 1/4.5
1/62=1/6.6 < 1/5.6
1/72=1/7.7 < 1/6.7
1/82=1/8.8 <1/7.8
⇒B<1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8
B<1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8
B<1/1-1/8
B<7/8
mà 7/8<1
⇒B<7/8<1
⇒B<1
b)S=3/1.4+3/4.7+3/7.10+...+3/40.43+3/43.46
S=1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46
S=1/1-1/46
S=45/46
Vì 45/46<1 nên S<1
Vậy S<1
Chúc bạn học tốt!
Bài 4:
$A+2=1+2+2^2+2^3+...+2^{11}$
$=(1+2)+(2^2+2^3)+....+(2^{10}+2^{11})$
$=(1+2)+2^2(1+2)+....+2^{10}(1+2)$
$=(1+2)(1+2^2+....+2^{10})$
$=3(1+2^2+...+2^{10})\vdots 3$
Vậy $A+2\vdots 3$ nên $A$ không chia hết cho $3$
Bài 5:
$n^2+n+1=n(n+1)+1$
Vì $n,n+1$ là hai số tự nhiên liên tiếp nên sẽ tồn tại một số chẵn và 1 số lẻ
$\Rightarrow n(n+1)$ chẵn
$\Rightarrow n^2+n+1=n(n+1)+1$ lẻ (điều phải chứng minh)
Lời giải:
a. Ta thấy:
$3+3^2+3^3+...+3^{99}\vdots 3$
$1\not\vdots 3$
$\Rightarrow A=1+3+3^2+...+3^{99}\not\vdots 3$
$\Rightarrow A\not\vdots 9$
b.
$A=(5+5^2)+(5^3+5^4)+...+(5^{39}+5^{40})$
$=5(1+5)+5^3(1+5)+...+5^{39}(1+5)$
$=5.6+5^3.6+....+5^{39}.6$
$=6(5+5^3+...+5^{39})$
$=2.3.(5+5^3+...+5^{39})$
$\Rightarrow A\vdots 2$ và $A\vdots 3$
dễ mà mình thi rồi giờ thì hết kiến thức ko làm đc sory ko làm đc
A=1/2*2+1/3*3+1/4*4+...+1/2017*2017.
=>A<1/1*2+1/25*3+1/3*4+...+1/2016*2017.
=>A<1-1/2+1/2-1/3+1/3-1/4+...+1/2016-1/2017.
=>A<1-1/2017<1.
=>A<1(đpcm).
tk mk nha có j kb.