Question

[Trần Thư]

1, P = \(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}\) - \(\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{2}{2004}}\)

2, Q = ( \(\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\) + \(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}\) ) : \(\dfrac{1980}{3758}\) + 155

3, A = 1.3 + 2.4 + 3.5 +....+ 97.99 + 98.100

4, B = 1.2.3 + 2.3.4. +...+ 48.49.50

5, C = \(\dfrac{1}{1.2.3.4}\) + \(\dfrac{1}{2.3.4.5}\) +...+ \(\dfrac{1}{27.28.29.30}\)

6, D = 1 + \(2^2\) + \(2^4\) + \(2^6\) + .... +\(2^{200}\)

7, E = \(\dfrac{1}{3.5}\)+ \(\dfrac{5}{5.7}\) +...+ \(\dfrac{1}{97.99}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

6:

\(4D=2^2+2^4+...+2^{202}\)

=>3D=2^202-1

hay \(D=\dfrac{2^{202}-1}{3}\)

7: \(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{32}{99}=\dfrac{16}{99}\)