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Ko bt đúng ko .
Đặt A=1.2+2.3+3.4+...+n(n+1)
A=1.2+2.3+3.4+...+n(n+1)
=>3A=(3−0)1.2+(4−1)2.3+...+(n+2−n+1)n(n+1)=>3A=(3−0)1.2+(4−1)2.3+...+(n+2−n+1)n(n+1)
=>3A=1.2.3−0.1.2+2.3.4−1.2.3+...+n(n+1)(n+2)−(n−1)n(n+1)=>3A=1.2.3−0.1.2+2.3.4−1.2.3+...+n(n+1)(n+2)−(n−1)n(n+1)
=>3A=n(n+1)(n+2)=>3A=n(n+1)(n+2)
=>A=n(n+1)(n+2)3=>A=n(n+1)(n+2)3 (đpcm)
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)
\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
\(=\frac{99}{100}-\frac{1}{2}\cdot\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)
A=1-1/2+1/2-1/3+...+1/n-1/n+1
=1-1/n+1
=n/n+1 không là số nguyên
\(A-1=\frac{1}{1.2}+\frac{1}{2.3}..+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}\)\(=\frac{99}{100}\)
\(A=1+\frac{99}{100}=\frac{199}{100}\)
1)
Để \(\left(2x^3-7x^2+5x+m\right)⋮\left(2x-3\right)\)
\(\Leftrightarrow m-\dfrac{3}{2}=0\\ \Leftrightarrow m=\dfrac{3}{2}\)
Vậy \(m=\dfrac{3}{2}\) thì \(\left(2x^3-7x^2+5x+m\right)⋮\left(2x-3\right)\)
\(1+\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=2-\frac{1}{100}\)
\(=\frac{199}{100}\)
Gọi biểu thức là A
A=1+1/2+1/2.3+1/3.4+...+1/98.99+1/99.100
A-1=1/2+1/2.3+1/3.4+...+1/98.99+1/99.100
A-1=1-1/2+1/2-1/3+1/3-1/4+...+/198-1/99+1/99-1/100
A-1=1-1/100
A-1=99/100
A=99/100+1
A=199/100
