সৃজনশীলঃ ‘p=a+bx2 একটি দ্বিপদী…’ যশোর বোর্ড ২০২৩ – দ্বিপদী বিস্তৃতি, উচ্চতর গণিত (এসএসসি)

দেওয়া আছে, $\mathrm{P}=\mathrm{a}+{\mathrm{bx}}^2$

যেখানে, $\mathrm{a}=\frac{1}{\mathrm{x}}$ এবং $\mathrm{b}=-2$

এখন, ${\mathrm{p}}^5={\mathrm{a}}^5+{}^5{\mathrm{C}}_1{\mathrm{a}}^4{\mathrm{bx}}^2+{}^5{\mathrm{C}}_2{\mathrm{a}}^3{\left({\mathrm{bx}}^2\right)}^2+{}^5{\mathrm{C}}_3{\mathrm{a}}^2{\left({\mathrm{bx}}^2\right)}^3$

$+{}^5{\mathrm{C}}_4\mathrm{a}{\left({\mathrm{bx}}^2\right)}^4+{}^5{\mathrm{C}}_5{\left({\mathrm{bx}}^2\right)}^5$

$=a^5+5a^{4}b{x}^2+10a^3{b}^2{x}^4+10a^2{b}^3{x}^6+5a{b}^4{x}^8+b^5{x}^{10}$

$=\frac{1}{{\mathrm{x}}^5}+5\times \frac{1}{{\mathrm{x}}^4}\times (-2)\times {\mathrm{x}}^2+10\times \frac{1}{{\mathrm{x}}^3}\times (-2{)}^2\times {\mathrm{x}}^4+10\times \frac{1}{{\mathrm{x}}^2}\times (-2{)}^3\times$

$x^6+5\times \frac{1}{x}\times (-2{)}^4\times x^8+(-2{)}^5{x}^{10}$

$\therefore {\mathrm{p}}^5=\frac{1}{{\mathrm{x}}^5}-\frac{10}{{\mathrm{x}}^2}+40\mathrm{x}-80{\mathrm{x}}^4+80{\mathrm{x}}^7-32{\mathrm{x}}^{10}$ (Ans.)

দেওয়া আছে, $p=a+b{x}^2$

বা, $p^8={(a+b{x}^2)}^8$

বা, $p^8={(2x^2+\frac{k}{x^3}{x}^2)}^8={(2x^2+\frac{k}{x})}^8$

$\therefore 8$ পথ বা $(3+1)$ তম পদ

$={}^8{\mathrm{C}}_3{\left(2{\mathrm{x}}^2\right)}^{8-3}{\left(\frac{\mathrm{k}}{\mathrm{x}}\right)}^3$

$={}^8{\mathrm{C}}_3\cdot 2^5\cdot {\mathrm{x}}^{10}\cdot {\mathrm{k}}^3\cdot {\mathrm{x}}^{-3}={}^8{\mathrm{C}}_3\cdot 2^5\cdot {\mathrm{k}}^3\cdot {\mathrm{x}}^7$

৫ম বা $(4+1)$ তম পদ

$={}^8{\mathrm{C}}_4{\left(2{\mathrm{x}}^2\right)}^{8-4}{\left(\frac{\mathrm{k}}{\mathrm{x}}\right)}^4$

$={}^8{\mathrm{C}}_4\cdot 2^4\cdot {\mathrm{x}}^8\cdot {\mathrm{k}}^4\cdot {\mathrm{x}}^{-4}={}^8{\mathrm{C}}_4\cdot 2^4\cdot {\mathrm{k}}^4\cdot {\mathrm{x}}^4$

প্রশ্নমতে, ${}^8{\mathrm{C}}_3{.2}^5.{\mathrm{k}}^3={}^8{\mathrm{C}}_4{.2}^4.{\mathrm{k}}^4$

বা, $\frac{{\mathrm{k}}^4}{{\mathrm{k}}^3}=\frac{{}^8{\mathrm{C}}_3{.2}^5}{{}^8{\mathrm{C}}_4{.2}^4}$

বা, $\mathrm{k}=\frac{56\times 32}{70\times 16}$

$\therefore \mathrm{k}=\frac{8}{5}$ (Ans.)