সৃজনশীলঃ ‘f(x)=c−bx−ax2;a≠0.…’ সরকারি হাজী মুহাম্মদ মহসিন কলেজ ২০২৬ – দ্বিপদী বিস্তৃতি, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $f\left(x\right)=c-\mathrm{bx}-{\mathrm{ax}}^2$

$f\left(x\right)=0$ হলে, $c-\mathrm{bx}-{\mathrm{ax}}^2=0$

বা, ${\mathrm{ax}}^2+\mathrm{bx}-c=0$
(i) নং সমীকরণের মূলদ্বয় $\alpha$ ও ${\alpha }^2$ হলে, $\alpha +{\alpha }^2=-\frac{b}{a}$ এবং $\alpha \cdot {\alpha }^2=-\frac{c}{a}$ বা, ${\alpha }^3=-\frac{c}{a}$

এখন, $\frac{(a-b{)}^3}{(c+b{)}^3}=\frac{a^3{(1-\frac{b}{a})}^3}{a^3{(\frac{c}{a}+\frac{b}{a})}^3}=\frac{{\{1+(-\frac{b}{a})\}}^3}{{\{-(-\frac{c}{a})-(-\frac{b}{a})\}}^3}$

$=\frac{{(1+\alpha +{\alpha }^2)}^3}{{\{-{\alpha }^3-(\alpha +{\alpha }^2)\}}^3}=\frac{{(1+\alpha +{\alpha }^2)}^3}{{(-\alpha -{\alpha }^2-{\alpha }^3)}^3}$

$=\frac{{(1+\alpha +{\alpha }^2)}^3}{{\{-\alpha (1+\alpha +{\alpha }^2)\}}^3}=\frac{{(1+\alpha +{\alpha }^2)}^3}{(-\alpha {)}^3{(1+\alpha +{\alpha }^2)}^3}$ 

$=\frac{1}{-{\alpha }^3}=\frac{1}{-(-\frac{c}{a})}=\frac{1}{a}=1\times \frac{a}{c}$

বা, $\frac{(a-b{)}^3}{(c+b{)}^3}=\frac{a}{c}$
$\therefore c(a-b{)}^3=a(c+b{)}^3$ (দেখানো হলো)

দেওয়া আছে,

$f\left(x\right)=c-bx-a{x}^2$
$a=2,b=5$ ও $c=3$ হলে, $f\left(x\right)=3-5x-2x^2$
$=3-6x+x-2x^2$
$=3(1-2x)+x(1-2x)$
$=(1-2x)(3+x)$
$\therefore \frac{1}{f\left(x\right)}=\frac{1}{(1-2x)(3+x)}$
$=\frac{\frac{2}{7}}{1-2x}+\frac{\frac{1}{7}}{3+x}$
$=\frac{2}{7}(1-2x{)}^{-1}+\frac{1}{7}(3+x{)}^{-1}$
$=\frac{2}{7}(1-2x{)}^{-1}+\frac{1}{21}{(1+\frac{x}{3})}^{-1}$
$=\frac{2}{7}\{1+2x+(2x{)}^2+(2x{)}^3+\dots .+(2x{)}^{n-1}\}+\frac{1}{21}\{1-\frac{x}{3}+{\left(\frac{x}{3}\right)}^2-{\left(\frac{x}{3}\right)}^3+\dots \dots +(-1{)}^{n-1}{\left(\frac{x}{3}\right)}^{n-1}\}$

$\therefore x^{n-1}$ এর সহগ $=\frac{2}{7}\cdot 2^{n-1}+(-1{)}^{n-1}\cdot \frac{1}{21}\cdot \frac{1}{3^{n-1}}$
$=\frac{2^{1+n-1}}{7}+(-1{)}^{n-1}\cdot \frac{1}{7\cdot 3^1\cdot 3^{n-1}}$
$=\frac{2^n}{7}+(-1{)}^{n-1}\frac{1}{7}\cdot \frac{1}{3^n}$

নির্ণেয় সহগ : $\frac{1}{7}\{2^n+(-1{)}^{n-1}\cdot \frac{1}{3^n}\}$