সৃজনশীলঃ ‘$f(x)=left(x^2+frac{3}{x}right)^{11}…………..(1)g(x)=(1+px)m …………..(2)…’ চট্টগ্রাম বোর্ড ২০১৭ – দ্বিপদী বিস্তৃতি, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

এখানে, $f\left(x\right)={(x^2+\frac{3}{x})}^{11}$

$f\left(x\right)={(x^2+\frac{3}{x})}^{11}$ এর বিস্তৃতিতে

(r+1) তম পদ

$\begin{array}{cc}& ={}^{11}{C}_r\cdot {\left(x^2\right)}^{11-r}\cdot {\left(\frac{3}{x}\right)}^r\\ & ={}^{11}{C}_r\cdot x^{22-2r}\cdot \frac{3^r}{x^r}={}^{11}{C}_r\cdot 3^r\cdot x^{22-3r}\end{array}$

(r+2) তম পদ

$\begin{array}{cc}& ={}^{11}{C}_{r+1}\cdot {\left(x^2\right)}^{11-r-1}\cdot {\left(\frac{3}{x}\right)}^{r+1}\\ & ={}^{11}{C}_{r+1}\cdot x^{22-2r-2}\cdot \frac{3^{r+1}}{x^{r+1}}\\ & ={}^{11}{C}_{r+1}\cdot 3^{r+1}\cdot x^{19-3r}\end{array}$

শর্তমতে, ${}^{11}{C}_r\cdot 3^r={}^{11}{C}_{r+1}\cdot 3^{r+1}$
বা, $\frac{11!}{r!(11-r)!}\cdot 3^r=\frac{11!}{(r+1)!(11-r-1)!}\cdot 3^r\cdot 3$
বা, $\frac{1}{r!(11-r)(10-r)!}=\frac{3}{(r+1)r!(10-r)!}$
বা, $\frac{1}{11-r}=\frac{3}{r+1}$
বা, $r+1=33-3r$
বা, $r+3r=33-1$
বা,  $4r=32$
বা,  $r=\frac{32}{4}$
বা, $r=8$
নির্ণেয় মান: : $r=8$

এখানে, $g\left(x\right)=(1+px{)}^m$

$p=-8$ এবং $m=-\frac{1}{2}$

$\therefore g\left(x\right)=(1-8x{)}^{-\frac{1}{2}}$ এর বিস্তৃতিতে $(r+1)$ তম পদ

$\begin{array}{cc}& =\frac{(-\frac{1}{2})(-\frac{1}{2}-1)(-\frac{1}{2}-2)\dots \dots (-\frac{1}{2}-r+1)}{r!}(-8x{)}^r\\ & =\frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})\dots \dots \cdot \{-\frac{1}{2}(2r-1\left)\right\}}{r!}(-8{)}^r{x}^r\\ & =\frac{1.3.5\dots \dots .(2r-1)(-\frac{1}{2})r}{r!}\cdot (-2.2.2{)}^r{x}^r\\ & =\frac{1.2.3\cdot 4.5.6\dots .(2r-1)\left(2r\right)\cdot \{(-\frac{1}{2})\cdot (-2\left)\right\}}{2.4.6\dots .\left(2r\right)\cdot r!}\cdot 2^r\cdot 2^r\cdot x^r\end{array}$

$\begin{array}{cc}& =-\frac{\left(2r\right)!\cdot 1^r\cdot 2^r\cdot 2^r}{(2\cdot 1)(2\cdot 2)(2\cdot 3)\dots \dots (2\cdot r)\cdot r!}{x}^r\\ & =\frac{\left(2r\right)!\cdot 2^r\cdot 2^r}{2^r\cdot (1\cdot 2\cdot 3\dots \dots \cdot r)\cdot r!}{x}^r=\frac{\left(2r\right)!\cdot 2^r\cdot 2^r}{2^r\cdot r!\cdot r!}{x}^r=\frac{\left(2r\right)!\cdot 2^r}{(r!{)}^2}{x}^r\end{array}$

$\therefore x^I$ এর সহগ $\frac{\left(2x\right)!\cdot 2^2}{\left(r\right){)}^2}$. (দেখানো হলো)