Bu muammo uchun haqiqatan ham unchalik ko'p narsa yo'q. Biz qilishimiz kerak bo'lgan yagona narsa, \(n = 2\) dan boshlab, ketma-ketlik shartlari formulasiga \(n\) ning dastlabki beshta qiymatini kiritish. Shunday qilib,

\[\begin{align*}& n = 2:\hspace{0.5in}\frac{{{{\left( { - 1} \right)}^{2 + 1}}}}{{2\left( 2 \right) + {{\left( { - 3} \right)}^2}}} = \frac{{ - 1}}{{13}} = - \frac{1}{{13}}\\ & n = 3:\hspace{0.5in}\frac{{{{\left( { - 1} \right)}^{3 + 1}}}}{{2\left( 3 \right) + {{\left( { - 3} \right)}^3}}} = \frac{1}{{ - 21}} = - \frac{1}{{21}}\\ & n = 4:\hspace{0.5in}\frac{{{{\left( { - 1} \right)}^{4 + 1}}}}{{2\left( 4 \right) + {{\left( { - 3} \right)}^4}}} = \frac{{ - 1}}{{89}} = - \frac{1}{{89}}\\ & n = 5:\hspace{0.5in}\frac{{{{\left( { - 1} \right)}^{5 + 1}}}}{{2\left( 5 \right) + {{\left( { - 3} \right)}^5}}} = \frac{1}{{ - 233}} = - \frac{1}{{233}}\\ & n = 6:\hspace{0.5in}\frac{{{{\left( { - 1} \right)}^{6 + 1}}}}{{2\left( 6 \right) + {{\left( { - 3} \right)}^6}}} = \frac{{ - 1}}{{741}} = - \frac{1}{{741}}\end{align*}\]

Shunday qilib, ketma-ketlikning birinchi beshta sharti:

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\left\{ { - \frac{1}{{13}},\,\, - \frac{1}{{21}},\,\, - \frac{1}{{89}},\,\, - \frac{1}{{233}},\,\, - \frac{1}{{741}},\,\, \ldots } \right\}}}\]

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.