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Bài 5:
b: Ta có: \(n+6⋮n+2\)
\(\Leftrightarrow n+2\in\left\{2;4\right\}\)
hay \(n\in\left\{0;2\right\}\)
c: Ta có: \(3n+1⋮n-2\)
\(\Leftrightarrow n-2\in\left\{-1;1;7\right\}\)
hay \(n\in\left\{1;3;9\right\}\)
a) \(1+2+3+4+...+n\)
\(=\left(n+1\right)\left[\left(n-1\right):1+1\right]:2\)
\(=\left(n+1\right)\left(n-1+1\right):2\)
\(=n\left(n+1\right):2\)
\(=\dfrac{n\left(n+1\right)}{2}\)
b) \(2+4+6+..+2n\)
\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)
\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)
\(=\left(n+1\right)\left(n-1+1\right)\)
\(=n\left(n+1\right)\)
c) \(1+3+5+...+\left(2n+1\right)\)
\(=\left[\left(2n+1\right)+1\right]\left\{\left[\left(2n-1\right)-1\right]:2+1\right\}:2\)
\(=\left(2n+1+1\right)\left[\left(2n-1-1\right):2+1\right]:2\)
\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)
\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)
\(=\left(n+1\right)\left(n-1+1\right)\)
\(=n\left(n+1\right)\)
d) \(1+4+7+10+...+2005\)
\(=\left(2005+1\right)\left[\left(2005-1\right):3+1\right]:2\)
\(=2006\cdot\left(2004:3+1\right):2\)
\(=2006\cdot\left(668+1\right):2\)
\(=1003\cdot669\)
\(=671007\)
e) \(2+5+8+...+2006\)
\(=\left(2006+2\right)\left[\left(2006-2\right):3+1\right]:2\)
\(=2008\cdot\left(2004:3+1\right):2\)
\(=1004\cdot\left(668+1\right)\)
\(=1004\cdot669\)
\(=671676\)
g) \(1+5+9+...+2001\)
\(=\left(2001+1\right)\left[\left(2001-1\right):4+1\right]:2\)
\(=2002\cdot\left(2000:4+1\right):2\)
\(=1001\cdot\left(500+1\right)\)
\(=1001\cdot501\)
\(=501501\)
a) 1+2+3+4+5+...+n = n(n+1) / 2
b)2+4+6+...+2n = [(2n-2):2+1] . (2n+2)/2 = n . ( 2n+2) /2
Cái tên.. àk mà thôi -_-
\(a)\) \(1+2+3+4+...+n=\frac{n\left(n+1\right)}{2}\)
\(b)\) \(2+4+6+8+...+2n=\left(\frac{2n-2}{2}+1\right)\left(2n+2\right)=\frac{2n\left(2n+2\right)}{2}=2n\left(n+1\right)\)
\(c)\) \(1+3+5+...+\left(2n+1\right)=\left(\frac{2n+1-1}{2}+1\right)\left(2n+1+1\right)=\frac{\left(2n+2\right)\left(2n+2\right)}{2}=\frac{\left(2n+2\right)^2}{2}\)
\(d)\) \(1+4+7+10+...+2005=\left(\frac{2005-1}{3}+1\right)\left(2005+1\right)=1342014\)
\(e)\) \(2+5+...+2006=\left(\frac{2006-2}{3}+1\right)\left(2006+2\right)=1343352\)
\(g)\) \(1+5+9+...+2001=\left(\frac{2001-1}{4}+1\right)\left(2001+1\right)=1003002\)
Chúc bạn học tốt ~
Bài 1:
\(\Leftrightarrow2^n\cdot\dfrac{9}{2}=9\cdot5^n\)
\(\Leftrightarrow2^n=2\cdot5^n\)
\(\Leftrightarrow2^{n-1}=5^n\)
Bài 2:
a: \(A=\dfrac{2^8+3^8}{2^8}=1+\dfrac{3^8}{2^8}\)
b: \(B=\left(2^{17}+17^2\right)\cdot\left(9^{15}-15^9\right)\cdot\left(16-16\right)=0\)
giúp mik với