10.18-bo'lim : binom seriyasi
4. \(\sqrt[3]{{8 - 2x}}\) uchun binom qatoridagi dastlabki to‘rtta hadni yozing.
Yechim
Birinchidan, binom seriyasidan foydalanish uchun uning to'g'ri shaklda ekanligiga ishonch hosil qilishimiz kerak. Mana bu funktsiya uchun to'g'ri shakl,
\[\sqrt[3]{{8 - 2x}} = {\left( {8\left( {1 - \frac{1}{4}x} \right)} \right)^{\frac{1}{3}}} = {\left( 8 \right)^{\frac{1}{3}}}{\left( {1 - \frac{1}{4}x} \right)^{\frac{1}{3}}} = 2{\left( {1 + \left( { - \frac{1}{4}x} \right)} \right)^{\frac{1}{3}}}\]Eslatib o'tamiz, to'g'ri kelishi uchun biz "1+" ko'rinishida bo'lishimiz kerak va shuning uchun biz ildizdan 8ni ajratib, minus belgisini ikkinchi hadga "ko'chirishimiz" kerak edi. Bundan tashqari, biz ko'rib turganimizdek, bizda \(k = \frac{1}{3}\) bo'ladi
2-qadam
Endi biz qilishimiz kerak bo'lgan yagona narsa eslatmalardan formulaga ulanish va dastlabki to'rtta shartni yozishdir.
\[\begin{align*}\sqrt[3]{{8 - 2x}} & = 2{\left( {1 + \left( { - \frac{1}{4}x} \right)} \right)^{\frac{1}{3}}}\\ & = 2\sum\limits_{i = 0}^\infty { {\frac{1}{3} \choose i} {{\left( { - \frac{1}{4}x} \right)}^i}} \\ & = 2\left[ {1 + \left( {\frac{1}{3}} \right){{\left( { - \frac{1}{4}x} \right)}^1} + \frac{{\left( {\frac{1}{3}} \right)\left( { - \frac{2}{3}} \right)}}{{2!}}{{\left( { - \frac{1}{4}x} \right)}^2} + \frac{{\left( {\frac{1}{3}} \right)\left( { - \frac{2}{3}} \right)\left( { - \frac{5}{3}} \right)}}{{3!}}{{\left( { - \frac{1}{4}x} \right)}^3} + \cdots } \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{2 - \frac{1}{6}x - \frac{1}{{72}}{x^2} - \frac{5}{{2592}}{x^3} + \cdots }}\end{align*}\]