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11 tháng 5 2016
\(\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{100}\left(1+2+...+100\right)\)
\(=\frac{1+2}{2}+\frac{1+2+3}{3}+...+\frac{1+2+...+100}{100}\)
\(=\frac{\left(1+2\right).2:2}{2}+\frac{\left(1+2+3\right).3:2}{3}+...+\frac{\left(1+2+...+100\right).100:2}{100}\)
\(=\left(1+2\right):2+\left(1+2+3\right):2+....\left(1+2+...+100\right):2\)
\(=\left[\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+...+100\right)\right]:2\)
\(=\left(100.1+99.2+....+1.100\right):2=171700:2=85850\)
Nếu không hiểu cái trong ngoặc tính sao thì báo tớ ;)
PD
0
24 tháng 10 2019
a) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n}\right)\\ =\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{n-1}{n}\\ =\frac{1}{n}\)
b) \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{n}\right)\\ =\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{n+1}{n}\\ =n+1\)
c) \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\\ =\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot\frac{3\cdot5}{4^2}\cdot...\cdot\frac{\left(n-1\right)\left(n+1\right)}{n^2}\\ =\frac{\left[1\cdot2\cdot3\cdot...\cdot\left(n-1\right)\right]\cdot\left[3\cdot4\cdot5\cdot...\cdot\left(n+1\right)\right]}{\left(2\cdot3\cdot4\cdot...\cdot n\right)\left(2\cdot3\cdot4\cdot...\cdot n\right)}\\ =\frac{n+1}{2n}\)
d) \(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)...\left(1+\frac{1}{99\cdot101}\right)\\ =\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot...\cdot\frac{10000}{99\cdot101}\\ =\frac{2^2\cdot3^2\cdot...\cdot100^2}{1\cdot3\cdot2\cdot4\cdot...\cdot99\cdot101}\\ =\frac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot99\right)\left(3\cdot4\cdot...\cdot101\right)}\\ =\frac{2\cdot100}{101}\\ =\frac{200}{101}\)
PT
1
19 tháng 4 2020
1 - 1/2 + 2 - 2/3 + 3 - 3/4 + 4 - 1/4 - 3 - 1/3 - 2 - 1/2 - 1
=(1-1)+(2-2)+(3-3)+\(\left(-\frac{1}{2}-\frac{1}{2}\right)+\left(-\frac{2}{3}-\frac{1}{3}\right)+\left(\frac{-3}{4}-\frac{1}{4}\right)\)+4
=0+(-3)+4
=1