Question

Tính tổng sau :

D = 1/1x3 + 1/3x5 + .... + 1/(2n-1)x(2n+1)

E = 1/1x3x5 + 1/3x5x7 + .... + 1/(2n-1)x(2n+1)x(2n+3)

mk cần gấp lắm

Answer 1

\(D= \dfrac{1}{1.3} + \dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right).\left(2n+1\right)}\),

\(2.D = \dfrac{2}{1.3}+ \dfrac{2}{3.5}+...+\dfrac{2}{\left(2n-1\right).\left(2n+1\right)}\)

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

\(2.D = 1 - \dfrac{1}{\left(2n+1\right)}\)

\(2.D= \dfrac{2n}{\left(2n+1\right)} \)

Vậy \(D = \dfrac{n}{\left(2n+1\right)}\)

Answer 2

\(E=\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+\dfrac{1}{5.7.9}+...+\dfrac{1}{\left(2n-1\right).\left(2n+1\right).\left(2n+3\right)}\)

\(\Rightarrow4E=4.\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+\dfrac{1}{5.7.9}+...+\dfrac{1}{\left(2n-1\right).\left(2n+1\right).\left(2n+3\right)}\)

\(=\dfrac{4}{1.3.5}+\dfrac{4}{3.5.7}+...+\dfrac{4}{\left(2n-1\right).\left(2n+1\right).\left(2n+3\right)}\)

\(=\dfrac{1}{1.3}-\dfrac{1}{3.5}+\dfrac{1}{3.5}-\dfrac{1}{5.7}-...+\dfrac{1}{\left(2n-1\right).\left(2n+1\right)}-\dfrac{1}{\left(2n+1\right).\left(2n+3\right)}\)

\(=\dfrac{1}{1.3}-\dfrac{1}{\left(2n+1\right).\left(2n+3\right)}\)

\(\Rightarrow E=\dfrac{\dfrac{1}{1.3}-\dfrac{1}{\left(2n+1\right).\left(2n+3\right)}}{4}\)

\(=\dfrac{1}{12}-\dfrac{1}{\left(2n+1\right).\left(2n+3\right).4}\)