tính giá trị của biểu thức
A=\(1+\frac{1}{2}\cdot\left(1+2\right)+\frac{1}{3}\cdot\left(1+2+3\right)+...+\frac{1}{100}\cdot\left(1+2+3+...+100\right)\)
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Ta có:
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right).\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{98}{99}.\frac{99}{100}\) \(=\frac{1.2.3...98.99}{2.3.4...99.100}=\frac{1}{100}\)
nha
B=\(\left(1-\dfrac{1}{1+2}\right)\). \(\left(1-\dfrac{1}{1+2+3}\right)\).....\(\left(1-\dfrac{1}{1+2+...+100}\right)\)
B=\(\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot...\cdot\left(1-\dfrac{1}{\left(1+100\right)\cdot100:2}\right)\)
B=\(\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{101\cdot100:2-1}{101\cdot100:2}\)
B=\(\dfrac{4}{6}\cdot\dfrac{10}{12}\cdot...\cdot\dfrac{\left(101.100:2-1\right).2}{101.100}\)
B=\(\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}\cdot...\cdot\dfrac{99.102}{100.101}\)
B=\(\dfrac{1.2.3.4.....99}{3.4.5....100}.\dfrac{4.5.6.....102}{3.4.5.....101}\)
B=\(\dfrac{2}{100}\).\(\dfrac{102}{3}\)
B=\(\dfrac{17}{25}\)
A=[(1+2+...+100) x (1/2 - 1/3 - 1/4 - 1/5) x (2,4x42 - 21x4,8)] / 1+1/2+1/3+...+1/100
= [(1+2+3+...+100) x (1/2 - 1/3 - 1/4-1/5) x (2,4x2x21 - 21x2x 4,8)] / 1+1/2+1/3+...+1/100
=[(1+2+3+...+100) x (1/2 - 1/3 - 1/4 - 1/5) x 0] / 1+1/2+1/3+...+1/100
=0 / 1+1/2+1/3+...+1/100 = 0
\(\frac{1}{2^2}-1=\frac{1-2^2}{2^2}=\frac{\left(1-2\right)\left(1+2\right)}{2^2}=-1.\frac{3}{2^2}\)
\(\frac{1}{3^2}-1=\frac{1-3^2}{3^2}=\frac{\left(1-3\right)\left(1+3\right)}{3^2}=-2.\frac{4}{3^2}\)
Đặt nguyên biểu thức là B , ta có :
\(B=\left[-1.\left(-2\right).\left(-3\right)...\left(-99\right)\right].\frac{3.4.5...101}{\left(2.3.4.5...100\right)^2}\)
\(B=-\left(1.2.3...99\right).\frac{3.4.5...101}{\left(2.3.4.5...100\right)^2}\)
B=\(\frac{-2.\left(3.4.5...99\right)^2.100.101}{2^2\left(3.4.5...99\right)^2.100^2}=\frac{-101}{200}\)