| Diketahui |
Diketahui 3 pernyataan yaitu:
- \({\rm{ }}{{\ddot a}_{\left. {\overline {\, n \,}}\! \right| }} = (1 + i){a_{\left. {\overline {\, n \,}}\! \right| }}\)
- \({{\ddot a}_{\left. {\overline {\, n \,}}\! \right| }} = {a_{\left. {\overline {\, n \,}}\! \right| }} – (1 + {v^n})\)
- \(\frac{1}{i} = \mathop {\lim }\limits_{n \to \infty } {a_{\left. {\overline {\, n \,}}\! \right| }}\)
|
| Rumus yang digunakan |
\({a_{\left. {\overline {\, n \,}}\! \right| }}_i = v + {v^2} + … + {v^n} = \frac{{1 – {v^n}}}{i}\)
\(\ddot a{{\rm{ }}_{\left. {\overline {\, n \,}}\! \right| }}_i = 1 + v + {v^2} + … + {v^{n – 1}} = \frac{{1 – {v^n}}}{d}\)
\(d = \frac{i}{{(1 + i)}}\)
\(v = \frac{1}{{(1 + i)}}\)
Perpetuity: \({a_{\left. {\overline {\, \infty \,}}\! \right| }}_i = \mathop {\lim }\limits_{n \to \infty } {a_{\left. {\overline {\, n \,}}\! \right| }}_i = v + {v^2} + … + {v^n} = \frac{1}{i}\) |
| Proses pengerjaan |
Pembuktian untuk pernyataan (1):
\(( \Leftarrow )(1 + i){a_{\left. {\overline {\, n \,}}\! \right| }} = (1 + i)\frac{{1 – {v^n}}}{i} = \frac{{(1 + i)}}{i}(1 – {v^n}) = \frac{{(1 – {v^n})}}{d} = \ddot a{{\rm{ }}_{\left. {\overline {\, n \,}}\! \right| }}_i\)
maka \(\ddot a{{\rm{ }}_{\left. {\overline {\, n \,}}\! \right| }}_i = (1 + i){a_{\left. {\overline {\, n \,}}\! \right| }}\)
Pembuktian untuk pernyataan (2):
\(( \Leftarrow ){a_{\left. {\overline {\, n \,}}\! \right| }}_i – (1 – {v^n}) = \frac{{1 – {v^n}}}{i} – (1 – {v^n}) = \frac{{1 – {v^n}}}{i} – \frac{{(1 – {v^n})i}}{i} = \frac{{(1 – i)(1 – {v^n})}}{i} \ne \ddot a{{\rm{ }}_{\left. {\overline {\, n \,}}\! \right| }}_i\)
maka \({\rm{ }}\ddot a{{\rm{ }}_{\left. {\overline {\, n \,}}\! \right| }}_i \ne {a_{\left. {\overline {\, n \,}}\! \right| }}_i – (1 – {v^n})\)
Pernyataan (3) sesuai dengan rumus Perpetuity dengan \(n \to \infty \) :
Perpetuity: \({a_{\left. {\overline {\, \infty \,}}\! \right| }}_i = \mathop {\lim }\limits_{n \to \infty } {a_{\left. {\overline {\, n \,}}\! \right| }}_i = v + {v^2} + … + {v^n} = \frac{1}{i}\) |
