Soal Eksponen (Pangkat)


Soal:

\frac{\frac{1}{{{1}^{4}}}+\frac{1}{{{2}^{4}}}+\frac{1}{{{3}^{4}}}+\frac{1}{{{4}^{4}}}+\frac{1}{{{5}^{4}}}+\frac{1}{{{6}^{4}}}+...}{\frac{1}{{{1}^{4}}}+\frac{1}{{{3}^{4}}}+\frac{1}{{{5}^{4}}}+\frac{1}{{{7}^{4}}}+...}=.......

Jawab:

Misalkan S=\frac{1}{{{1}^{4}}}+\frac{1}{{{2}^{4}}}+\frac{1}{{{3}^{4}}}+\frac{1}{{{4}^{4}}}+\frac{1}{{{5}^{4}}}+\frac{1}{{{6}^{4}}}+\frac{1}{{{7}^{4}}}+...

dan A=\frac{1}{{{1}^{4}}}+\frac{1}{{{3}^{4}}}+\frac{1}{{{5}^{4}}}+...

Berarti
\begin{array}{l}S=\frac{1}{{{1}^{4}}}+\frac{1}{{{3}^{4}}}+\frac{1}{{{5}^{4}}}+...+\frac{1}{{{2}^{4}}}+\frac{1}{{{4}^{4}}}+\frac{1}{{{6}^{4}}}+...\\\Leftrightarrow S=A+\frac{1}{{{(1\cdot 2)}^{4}}}+\frac{1}{{{(2\cdot 2)}^{4}}}+\frac{1}{{{(3\cdot 2)}^{4}}}+...\\\Leftrightarrow S=A+\frac{1}{{{1}^{4}}\cdot {{2}^{4}}}+\frac{1}{{{2}^{4}}\cdot {{2}^{4}}}+\frac{1}{{{3}^{4}}\cdot {{2}^{4}}}+...\\\Leftrightarrow S=A+\frac{1}{{{2}^{4}}}\left( \frac{1}{{{1}^{4}}}+\frac{1}{{{2}^{4}}}+\frac{1}{{{3}^{4}}}+... \right)\\\Leftrightarrow S=A+\frac{1}{16}S\\\Leftrightarrow A=S-\frac{1}{16}S=\frac{15}{16}S\end{array}

Substitusi A=\frac{15}{16}S ke soal, diperoleh

\frac{\frac{1}{{{1}^{4}}}+\frac{1}{{{2}^{4}}}+\frac{1}{{{3}^{4}}}+\frac{1}{{{4}^{4}}}+\frac{1}{{{5}^{4}}}+\frac{1}{{{6}^{4}}}+...}{\frac{1}{{{1}^{4}}}+\frac{1}{{{3}^{4}}}+\frac{1}{{{5}^{4}}}+\frac{1}{{{7}^{4}}}+...}=\frac{S}{A}=\frac{S}{\frac{15}{16}S}=\frac{16}{15}\,\,

Sekian. Semoga berguna.

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