C=(1-4/1)(1-4/9)(1-4/25)...(1-4/(2n-1)2 với n>=1
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BH
1
LQ
0
13 tháng 6 2016
\(A=\left(\frac{1^2-2^2}{1^2}\right)\left(\frac{3^2-2^2}{3^2}\right)\left(\frac{5^2-2^2}{5^2}\right)...\left(\frac{\left(2n-1\right)^2-2^2}{\left(2n-1\right)^2}\right)\)
\(=\frac{-1\cdot3}{1^2}\cdot\frac{1\cdot5}{3^2}\cdot\frac{3\cdot7}{5^2}...\cdot\frac{\left(2n-3\right)\left(2n+1\right)}{\left(2n-1\right)^2}=-\frac{1}{1}\cdot\frac{2n+1}{2n-1}=-\frac{2n+1}{2n-1}\)
H9
HT.Phong (9A5)
CTVHS
15 tháng 8 2023
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) \(1+3+5+...+\left(2n+1\right)\)
\(=\left[\left(2n+1\right)+1\right]\left\{\left[\left(2n-1\right)-1\right]:2+1\right\}:2\)
\(=\left(2n+1+1\right)\left[\left(2n-1-1\right):2+1\right]:2\)
\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)
\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)
\(=\left(n+1\right)\left(n-1+1\right)\)
\(=n\left(n+1\right)\)
H9
HT.Phong (9A5)
CTVHS
15 tháng 8 2023
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) \(1+5+9+...+2001\)
\(=\left(2001+1\right)\left[\left(2001-1\right):4+1\right]:2\)
\(=2002\cdot\left(2000:4+1\right):2\)
\(=1001\cdot\left(500+1\right)\)
\(=1001\cdot501\)
\(=501501\)
MN
0