সৃজনশীলঃ ‘A(2+x3)n এবং B=(1−x)(1+ax)5…’ বরিশাল বোর্ড ২০২৩ – দ্বিপদী বিস্তৃতি, উচ্চতর গণিত (এসএসসি)

$A={(2+\frac{x}{3})}^n$ বিস্তৃতির, $r+1$ তম পদ $={}^n{C}_r{2}^{n-r}\cdot {\left(\frac{x}{3}\right)}^r$

$\therefore (4+1)$ তম পদ $={}^n{C}_4{2}^{n-4}\cdot {\left(\frac{x}{3}\right)}^4$

$(5+1)$ তম পদ $={}^n{C}_5{2}^{n-5}\cdot {\left(\frac{x}{3}\right)}^5$

প্রশ্নমতে, ${}^n{C}_4{2}^{n-4}{\left(\frac{1}{3}\right)}^4=5\times {}^n{C}_5{2}^{n-5}{\left(\frac{1}{3}\right)}^5$

বা, $\frac{n!}{4!(n-4)!}\times 2^{n-4-n+5}\times 3^{-4+5}=5\times \frac{n!}{5!(n-5)!}$

বা, $\frac{2\times 3}{4!(n-4)(n-5)!}=5\times \frac{1}{5\times 4!(n-5)!}$

বা, $\frac{6}{n-4}=1$

বা, $n-4=6$

$\therefore n=10$ (Ans.)

দেওয়া আছে, $B=(1-x)(1+\mathrm{ax}{)}^5$

$=(1-x)\{1+{}^5{C}_1(\mathrm{ax}{)}^1+{}^5{C}_2(\mathrm{ax}{)}^2+{}^5{C}_3(\mathrm{ax}{)}^3+\dots \dots \}$
$=(1-x)(1+5\mathrm{ax}+10a^2{x}^2+10a^3{x}^3+\dots \dots )$
$=(1+5\mathrm{ax}+10a^2{x}^2+10a^3{x}^3+\dots \dots )$

$-(x+5{\mathrm{ax}}^2+10a^2{x}^3+10a^3{x}^4+\dots \dots )$

$=1+x(5a-1)+x^2(10a^2-5a)+x^3(10a^3-10a^2)+\dots \dots \left(i\right)$

প্রশ্নমতে, $B=1+{\mathrm{bx}}^2+{\mathrm{cx}}^3+$. ........ (ii)

(i)নং ও (ii)নং এ সহগ সমীকৃত করে পাই, $5a-1=0$

$\therefore a=\frac{1}{5}$

আবার, $10a^2-5a=b$

বা, $10\times {\left(\frac{1}{5}\right)}^2-5\cdot \frac{1}{5}=b$ [a এর মান বসিয়ে]

বা, $10\times \frac{1}{25}-1=b$

$\therefore b=-\frac{3}{5}$

এবং $10a^3-10a^2=c$

বা, $10\times {\left(\frac{1}{5}\right)}^3-10{\left(\frac{1}{5}\right)}^2=c$

বা, $\frac{10}{125}-\frac{10}{25}=c\therefore c=\frac{-8}{25}$

$\therefore a,b,c$ এর মান যথাক্রমে $\frac{1}{5},-\frac{3}{5},-\frac{8}{25}$. (Ans.)