সৃজনশীলঃ ‘দৃশ্যকল্প-১:f(x)=x2+14×2−1দৃশ্যকল্প-২: g(x)= 11−6x.…’ ভিকারুননিসা নূন স্কুল এন্ড কলেজ ২০২৬ – দ্বিপদী বিস্তৃতি, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $f\left(x\right)=x^2+\frac{1}{4x^2}-1$

$=x^2-2\cdot x\cdot \frac{1}{2x}+{\left(\frac{1}{2x}\right)}^2={(x-\frac{1}{2x})}^2$
$\therefore \left\{f\right(x){\}}^n={\left\{{(x-\frac{1}{2x})}^2\right\}}^n={(x-\frac{1}{2x})}^{2n}$
${(x-\frac{1}{2x})}^{2n}$ এর ঘাত $2n$ যা একটি জোড় সংখ্যা

∴ মধ্যপদটি হবে $(\frac{2n}{2}+1)$ তম পদ বা $(n+1)$ তম পদ

$\therefore (n+1)$ তম পদ $={}^{2n}{C}_n\cdot x^{2n-n}\cdot {(-\frac{1}{2x})}^n$
$=\frac{\left(2n\right)!}{(2n-n)!n!}\cdot x^n\cdot \frac{1}{2^n\cdot x^n}\cdot (-1{)}^n$
$=\frac{\left(2n\right)!}{n!n!2^n}(-1{)}^n$
$=\frac{1.2.3\cdot \dots \dots (2n-1)2n}{n!n!2^n}(-1{)}^n$
$=\frac{\{2\cdot 4\cdot 6\dots (2n-2\left)2n\right\}\{1.3\cdot 5\dots \dots (2n-1\left)\right\}}{n!n!2^n}(-1{)}^n$
${=\frac{2^n\{1\cdot 2\cdot 3\dots \dots (n-1)\cdot x\}\{1\cdot 3\cdot 5\dots \dots (2n-1\left)\right\}}{n!n!2^n})-1)}^n$
$=\frac{n!\{1.3.5\dots \dots (2n-1\left)\right\}}{n!n!}(-1{)}^n$
$=\frac{1.3.5\dots \dots (2n-1)}{n!}(-1{)}^n$

$\therefore \left\{f\right(x){\}}^n$ এর বিসৃতিতে $(r+1)$ তম পদের সহগ

$\frac{1.3.5\dots \dots .(2n-1)}{n!}(-1{)}^n$ (দেখানো হলো )

দেওয়া আছে, $g\left(x\right)=\frac{1}{\sqrt{1-6x}}=\frac{1}{(1-6x{)}^{\frac{1}{2}}}=(1-6x{)}^{-\frac{1}{2}}$
$(1-6x{)}^{-\frac{1}{2}}$ এর বিস্তৃতিতে $(r+1)$ তম পদ

$=\frac{-\frac{1}{2}\cdot (-\frac{1}{2}-1)(-\frac{1}{2}-2)\dots (-\frac{1}{2}-r+1)}{r!}(-6x{)}^r$
$=\frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})\dots \dots \{-\frac{1}{2}(2r-1\left)\right\}}{r!}(-6x{)}^r$
$=\frac{1.3\cdot 5\dots ..(2r-1){(-\frac{1}{2})}^r}{r!}(-2{)}^r{3}^r\cdot x^r$
$=\frac{1.3\cdot 5\dots \dots (2r-1)}{r!}{3}^r\cdot x^r$
$=\frac{\{1.2\cdot 5\cdot 7\dots \dots \dots (2r-1\left)\right\}\{2.4\cdot 6.8\dots \dots \dots .2r\}}{r!(2.4\cdot 8.8\dots \dots 2r)}{3}^r{x}^r$
$=\frac{1.2\cdot 3.4\dots \dots .2r}{r!2^r(1.2\cdot 3\dots \dots .r)}\cdot 3^r\cdot x^2$
$=\frac{\left(2r\right)!3^r}{2^r\cdot r!r!}{x}^r=\frac{\left(2r\right)!3^r}{2^r(r!{)}^2}{x}^r$
$\therefore x^r$ এর সহগ $\frac{\left(2r\right)!3^r}{2^r(r!{)}^2}$ (দেখানো হলো)