Question

S=1 pe 1x3 + 1 pe 3x5 + 1 pe 5x7+...+ 1 pe 89x91

Answer 1

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Answer 2

[tex]\it \dfrac{1}{k}-\dfrac{1}{k+2} = \dfrac{k+2-k}{k(k+2)} = \dfrac{2}{k(k+2)} \Longrightarrow \dfrac{1}{k(k+2)} = \dfrac{1}{2}\left( \dfrac{1}{k}-\dfrac{1}{k+2}\right) \\ \\ \\ S = \dfrac{1}{2}\left(1- \dfrac{1}{3} + \dfrac{1}{3} -\dfrac{1}{5} +\ ...\ +\dfrac{1}{89}-\dfrac{1}{91}\right) =\dfrac{1}{2}\left(1-\dfrac{1}{91}\right) = \\ \\ \\ = \dfrac{1}{2}\cdot\dfrac{90}{91} = \dfrac{45}{91}[/tex]