Keyinchalik, biz maxraj \(1 - p\) ko'rinishida bo'lishi kerakligini bilamiz va yana bu juda oson, bu holda maxrajdan 5 ni koeffitsientga ajratib, maxrajni quyidagicha qayta yozish mumkin,

\[f\left( x \right) = \frac{{3{x^2}}}{5}\frac{1}{{1 - \frac{2}{5}\,\sqrt[3]{x}}}\]

Ushbu nuqtada biz buni kuch seriyasi sifatida yozish uchun eslatmalardan formuladan foydalanishimiz mumkin. Shunday qilib,

\[f\left( x \right) = \frac{{3{x^2}}}{5}\frac{1}{{1 - \frac{2}{5}\,\sqrt[3]{x}}} = \frac{{3{x^2}}}{5}\sum\limits_{n = 0}^\infty {{{\left( {\frac{2}{5}\,\sqrt[3]{x}} \right)}^n}} \hspace{0.25in}\hspace{0.25in}{\mbox{provided}}\,\,\left| {\frac{2}{5}\,\sqrt[3]{x}} \right| < 1\]

Endi ketma-ket javob shakli uchun asosiy "qoidalarni" eslang. Biz seriyalar oldida hech narsa bo'lishini xohlamaymiz va biz bitta ko'rsatkichli bitta \(x\) ni xohlaymiz.

Bu g'amxo'rlik qilish uchun etarlicha oson qoidalar. Biz qilishimiz kerak bo'lgan narsa, seriya oldidagi narsalarni seriyaning ichki qismiga o'tkazish va \(x\) "qoidasi" ga rioya qilish uchun asosiy ko'rsatkich qoidalaridan foydalanishdir. Bularning barchasini qilish,

\[f\left( x \right) = \frac{{3{x^2}}}{5}\sum\limits_{n = 0}^\infty {{{\left( {\frac{2}{5}\,\sqrt[3]{x}} \right)}^n}} = \sum\limits_{n = 0}^\infty {\frac{3}{5}{x^2}{{\left( {\frac{2}{5}} \right)}^n}{{\left( {{x^{\frac{1}{3}}}} \right)}^n}} = \sum\limits_{n = 0}^\infty {\frac{3}{5}{{\left( {\frac{2}{5}} \right)}^n}{x^{\frac{1}{3}\,n + 2}}} \hspace{0.25in}\hspace{0.25in}{\mbox{provided}}\,\,\left| {\frac{2}{5}\,\sqrt[3]{x}} \right| < 1\]

Konvergentsiya oralig'ini olish uchun biz qilishimiz kerak bo'lgan yagona narsa natijaning "ta'minlangan" qismiga erishish uchun oxirgi bosqichdan boshlab bir oz ish qilishdir,

\[\left| {\frac{2}{5}\,\sqrt[3]{x}} \right| < 1\,\,\,\, \to \,\,\,\,\,\frac{2}{5}{\left| x \right|^{\frac{1}{3}}} < 1\hspace{0.25in} \to \hspace{0.25in}{\left| x \right|^{\frac{1}{3}}} < \frac{5}{2}\hspace{0.25in} \to \hspace{0.25in}\left| x \right| < \frac{{125}}{8}\hspace{0.25in} \to \hspace{0.25in} - \frac{{125}}{8}\, < x < \frac{{125}}{8}\]

E'tibor bering, biz ushbu oraliqning so'nggi nuqtalarini tekshirishimiz shart emas, chunki biz faqat qat'iy tengsizliklar bilan yaqinlashishimizni bilamiz va qolgan hamma narsa uchun farq qilamiz.

Bu muammoning javoblari shunday bo'ladi:

\[\require{bbox} \bbox[2pt,border:1px solid black]{{{\mbox{Power Series : }}\frac{{3{x^2}}}{{5 - 2\,\sqrt[3]{x}}} = \sum\limits_{n = 0}^\infty {\frac{3}{5}{{\left( {\frac{2}{5}} \right)}^n}{x^{\frac{1}{3}\,n + 2}}} }}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{{\mbox{Interval : }} - \frac{{125}}{8}\, < x < \frac{{125}}{8}\,}}\]