10.5-bo'lim : Maxsus seriyalar
1. Qatlamning yaqinlashish yoki ayrilishligini aniqlang. Agar qator yaqinlashsa, uning qiymatini bering.
\[\sum\limits_{n = 0}^\infty {{3^{2 + n}}\,{2^{1 - 3n}}} \]Yechim
Ushbu bo'limda biz ko'rib chiqqan uchta maxsus seriyaning barchasi bir-biridan farqli ekanligini hisobga olsak, bu geometrik qator ekanligi aniq bo'ladi.
2-qadam
Keling, indeksning boshlang'ich qiymati \(n = 0\) ekanligini va shuning uchun biz buni shaklga qo'yishimiz mumkinligiga e'tibor bering,
\[\sum\limits_{n = 0}^\infty {a\,{r^n}} \]O'sha paytda biz uning yaqinlashishi yoki uzoqlashishini va agar u yaqinlashayotgan bo'lsa, seriyaning qiymatini aniqlashimiz mumkin.
Bunday holda, seriyani yuqoridagi shaklga qo'yish juda oddiy, shuning uchun bu ish.
\[\sum\limits_{n = 0}^\infty {{3^{2 + n}}\,{2^{1 - 3n}}} = \sum\limits_{n = 0}^\infty {{3^2}{3^n}\,{2^1}{2^{ - 3n}}} = \sum\limits_{n = 0}^\infty {\left( 9 \right)\left( 2 \right)\frac{{{3^n}}}{{{2^{3n}}}}} = \sum\limits_{n = 0}^\infty {18\frac{{{3^n}}}{{{8^n}}}} = \sum\limits_{n = 0}^\infty {18{{\left( {\frac{3}{8}} \right)}^n}} \]Shartlarda bo'lishi mumkin bo'lgan har qanday salbiy ko'rsatkichlar bilan to'g'ri muomala qilganingizga ishonch hosil qiling!
Shuni ham yodda tutingki, barcha ko'rsatkichlar oddiy \(n\) bo'lishi kerak va 3\(n\) yoki boshqa narsa bo'lishi mumkin emas. Shunday qilib, bu muammo uchun biz \({2^{3n}} = {\left( {{2^3}} \right)^n} = {8^n}\) yozish uchun asosiy ko'rsatkich qoidalaridan foydalanishimiz kerak.
3-qadam
"To'g'ri" shakldagi seriyalar bilan biz \(a = 18\) va \(r = \frac{3}{8}\) ekanligini ko'rishimiz mumkin. Shuning uchun, biz aniq \(\left| r \right| = \frac{3}{8} < 1\) ko'rish mumkin, chunki, qator bo'ladi birlashish va uning qiymati,
\[\sum\limits_{n = 0}^\infty {{3^{2 + n}}\,{2^{1 - 3n}}} = \sum\limits_{n = 0}^\infty {18{{\left( {\frac{3}{8}} \right)}^n}} = \frac{{18}}{{1 - \frac{3}{8}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{144}}{5}}}\]