\(\frac{1^{2n-1}}{2}=\frac{1}{8}\)
\(1^{2n-1}=1\cdot2:8\)
\(1^{2n-1}=\frac{1}{4}\) ( vô lí vì \(1^{2n-1}=1\forall n\)
Vậy không có n thỏa mãn
\(\frac{1^{2n-1}}{2}=\frac{1}{8}\)
\(\Leftrightarrow\frac{4.\left(1^{2n-1}\right)}{8}=\frac{1}{8}\)
\(\Leftrightarrow1^{2n-1}=\frac{1}{4}\)
\(\Leftrightarrow1^{2n}=\frac{1}{4}\)
\(\Leftrightarrow1^n.1^2=\frac{1}{4}\)
\(\Leftrightarrow n=-4\)
Ta có: \(\frac{1^{2n-1}}{2}=\frac{1}{8}\)
Vì \(1^{2n-1}=1\)\(\Rightarrow\)\(\frac{1^{2n-1}}{2}=\frac{1}{2}\)mà \(\frac{1^{2n-1}}{2}=\frac{1}{8}\)
\(\Rightarrow\)\(S=\varnothing\)
