Question

1/(2.4) + 1/(4.6) + … + 1/[(2x – 2).2x] = 1/8 (x ∈ N, x ≥ 2)

giup vs

Answer 1

\(A=\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right).2x}=\dfrac{1}{8}\)

\(2A=\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right).2x}=\dfrac{1}{8}.2=\dfrac{2}{8}=\dfrac{1}{4}\)

\(2A=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{1}{4}\)

\(2A=\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{1}{4}\)

\(A=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{2x}\right)=\dfrac{1}{8}\)

\(A=\dfrac{1}{2}.\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{2x}=\dfrac{1}{8}\)

\(A=\dfrac{1}{4}-\dfrac{1}{4x}=\dfrac{1}{8}\)

\(\Rightarrow\dfrac{1}{4x}=\dfrac{1}{4}-\dfrac{1}{8}=\dfrac{2}{8}-\dfrac{1}{8}=\dfrac{1}{8}\)

\(\Rightarrow4x=8\)

\(\Rightarrow x=8:4=2\)

Answer 2

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