Bu muammo uchun haqiqatan ham unchalik ko'p narsa yo'q. Biz qilishimiz kerak bo'lgan yagona narsa, \(n = 0\) dan boshlab, ketma-ketlik shartlari formulasiga \(n\) ning dastlabki beshta qiymatini kiritish. Shunday qilib,

\[\begin{align*} & n = 0:\hspace{0.5in}\frac{{4\left( 0 \right)}}{{{{\left( 0 \right)}^2} - 7}} = 0\\ & n = 1:\hspace{0.5in}\frac{{4\left( 1 \right)}}{{{{\left( 1 \right)}^2} - 7}} = \frac{4}{{ - 6}} = - \frac{2}{3}\\ & n = 2:\hspace{0.5in}\frac{{4\left( 2 \right)}}{{{{\left( 2 \right)}^2} - 7}} = \frac{8}{{ - 3}} = - \frac{8}{3}\\ & n = 3:\hspace{0.5in}\frac{{4\left( 3 \right)}}{{{{\left( 3 \right)}^2} - 7}} = \frac{{12}}{2} = 6\\ & n = 4:\hspace{0.5in}\frac{{4\left( 4 \right)}}{{{{\left( 4 \right)}^2} - 7}} = \frac{{16}}{9}\end{align*}\]

Shunday qilib, ketma-ketlikning birinchi beshta sharti:

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\left\{ {0,\,\, - \frac{2}{3},\,\, - \frac{8}{3},\,\,6,\,\,\frac{{16}}{9},\,\, \ldots } \right\}}}\]

E'tibor bering, biz ketma-ketlik bilan shug'ullanayotganimizni unutmasligimiz uchun rasmiy javobni qavslar ichiga qo'yamiz va bu ketma-ketlikda biz sanab o'tgan beshtadan ko'ra ko'proq atamalar mavjudligini eslatish uchun oxiriga "..." ni qo'shdik.