D=2/3+3/32+4/33+...+101/3100
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5 tháng 11 2021
\(\Leftrightarrow3B=3^2+3^3+...+3^{101}\\ \Leftrightarrow3B-B=3^{101}-3\\ \Leftrightarrow2B=3^{101}-3\\ \Leftrightarrow2B+3=3^{101}=3^n\\ \Leftrightarrow n=101\)
24 tháng 12 2021
\(a,A=3+3^2+3^3+3^4+...+3^{100}\\ 3A=3^2+3^3+3^4+3^5+3^{101}\\ 3A-A=2A=3^{101}-3\\ \Rightarrow2A+3=3^{101}=3^{4.25+1}\\ \Rightarrow n=25\)
H9
HT.Phong (9A5)
CTVHS
11 tháng 1 2024
Câu 3:
\(A=3+3^2+...+3^{100}\)
\(3A=3^2+3^3+...+3^{101}\)
\(3A-A=3^2+3^3+...+3^{101}-\left(3+3^2+...+3^{100}\right)\)
\(2A=3^{101}-3\)
Mà: \(2A+3=3^N\)
\(\Rightarrow3^{101}-3+3=3^N\)
\(\Rightarrow3^{101}=3^N\)
\(\Rightarrow N=101\)
Vậy: ...
11 tháng 1 2024
Câu 1:
\(A=4+2^2+...+2^{20}\)
Đặt \(B=2^2+2^3+...+2^{20}\)
=>\(2B=2^3+2^4+...+2^{21}\)
=>\(2B-B=2^3+2^4+...+2^{21}-2^2-2^3-...-2^{20}\)
=>\(B=2^{21}-4\)
=>\(A=B+4=2^{21}-4+4=2^{21}\) là lũy thừa của 2
Câu 6:
Đặt A=1+2+3+...+n
Số số hạng là \(\dfrac{n-1}{1}+1=n-1+1=n\left(số\right)\)
=>\(A=\dfrac{n\left(n+1\right)}{2}\)
=>\(A⋮n+1\)
Câu 5:
\(A=5+5^2+...+5^8\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\left(5^5+5^6\right)+\left(5^7+5^8\right)\)
\(=\left(5+5^2\right)+5^2\left(5+5^2\right)+5^4\left(5+5^2\right)+5^6\left(5+5^2\right)\)
\(=30\left(1+5^2+5^4+5^6\right)⋮30\)
26 tháng 6 2022
a: \(=3\cdot\left(\dfrac{1}{4}-\dfrac{6}{7}+\dfrac{8}{21}\right)\)
\(=3\cdot\left(\dfrac{21}{84}-\dfrac{72}{84}+\dfrac{32}{84}\right)\)
\(=\dfrac{-19}{28}\)
b: \(=\dfrac{-2}{3}\left(\dfrac{1}{9}-\dfrac{1}{6}-\dfrac{1}{11}\right)\)
\(=\dfrac{-2}{3}\cdot\dfrac{-29}{198}=\dfrac{29}{99\cdot3}=\dfrac{29}{297}\)
c: \(=\dfrac{-3}{7}+\dfrac{4}{25}+\dfrac{5}{16}+\dfrac{3}{16}\)
\(=\dfrac{-75+28}{175}+\dfrac{1}{2}\)
\(=\dfrac{-47}{175}+\dfrac{1}{2}=\dfrac{-94+175}{350}=\dfrac{81}{350}\)
d: \(=\dfrac{-4}{9}\cdot\left(\dfrac{1}{26}-\dfrac{1}{2}-\dfrac{1}{8}\right)\)
\(=\dfrac{-4}{9}\cdot\dfrac{-61}{104}=\dfrac{61}{26\cdot9}=\dfrac{61}{234}\)
KV
23 tháng 12 2023
A = 8⁸ + 2²⁰
= (2³)⁸ + 2²⁰
= 2²⁴ + 2²⁰
= 2²⁰.(2⁴ + 1)
= 2²⁰.17 ⋮ 17
Vậy A ⋮ 17
20 tháng 11 2023
a: (x-3)(y+1)=15
=>\(\left(x-3\right)\left(y+1\right)=1\cdot15=15\cdot1=\left(-1\right)\cdot\left(-15\right)=\left(-15\right)\cdot\left(-1\right)=3\cdot5=5\cdot3=\left(-3\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-3\right)\)
=>(x-3;y+1)\(\in\){(1;15);(15;1);(-1;-15);(-15;-1);(3;5);(5;3);(-3;-5);(-5;-3)}
=>(x,y)\(\in\){(4;14);(18;0);(2;-16);(-12;-2);(6;4);(8;2);(0;-6);(-2;-4)}
b: Sửa đề:\(m=1+3+3^2+3^3+...+3^{99}+3^{100}\)
\(m=1+3+\left(3^2+3^3+3^4\right)+\left(3^5+3^6+3^7\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)
\(=4+3^2\left(1+3+3^2\right)+3^5\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)
\(=4+13\left(3^2+3^5+...+3^{98}\right)\)
=>m chia 13 dư 4
\(m=1+3+3^2+...+3^{99}+3^{100}\)
\(=1+\left(3+3^2+3^3+3^4\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(=1+3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+...+3^{97}\left(1+3+3^2+3^3\right)\)
\(=1+40\left(3+3^5+...+3^{97}\right)\)
=>m chia 40 dư 1
\(D=\frac{2}{3}+\frac{3}{3^2}+...+\frac{101}{3^{100}}\)
\(\Rightarrow3D=2+\frac{3}{3}+...+\frac{101}{3^{99}}\)
\(\Rightarrow2D=\left(2-\frac{101}{3^{100}}\right)+\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)\)
Đặt \(E=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(\Rightarrow3E=1+\frac{1}{3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow2E=1-\frac{1}{3^{99}}\Rightarrow E=\frac{1-\frac{1}{3^{99}}}{2}\)
\(\Rightarrow2D=\left(2-\frac{101}{3^{100}}\right)+\left(\frac{1-\frac{1}{3^{99}}}{2}\right)\)
\(\Rightarrow D=\frac{\left(2-\frac{101}{3^{100}}\right)+\left(\frac{1-\frac{1}{3^{99}}}{2}\right)}{2}\)