সৃজনশীলঃ ‘(i) g(y)=2y31−y3(ii) B=(2+x5)n;n∈N…’ দিনাজপুর বোর্ড ২০২৫ – দ্বিপদী বিস্তৃতি, উচ্চতর গণিত (এসএসসি)

দেওয়া আছে, $B={(2+\frac{x}{5})}^n;n\in N$

দ্বিপদী বিস্তৃতির সাহায্যে পাই,

$(2+\frac{x}{5})n=(\frac{n}{0})2^n{\left(\frac{x}{5}\right)}^0+(\frac{n}{1})2^{n-1}\left(\frac{x}{5}\right)+(\frac{n}{2})2^{n-2}$

${\left(\frac{x}{5}\right)}^2+(\frac{n}{3})2^{n-3}{\left(\frac{x}{5}\right)}^3+(\frac{n}{4})2^{n-4}{\left(\frac{x}{5}\right)}^4+(\frac{n}{2})2^{n-5}{\left(\frac{x}{5}\right)}^5$

$+(\frac{n}{6})2^{n-6}{\left(\frac{x}{5}\right)}^6+(\frac{n}{7})2^{n-7}{\left(\frac{x}{5}\right)}^7+(\frac{n}{8})2^{n-8}{\left(\frac{x}{5}\right)}^8$

$+(\frac{n}{9})2^{n-9}{\left(\frac{x}{5}\right)}^9+\dots \dots \dots$

$\therefore x^8$ এর সহগ $=(\frac{n}{8})2^{n-8}{\left(\frac{1}{5}\right)}^8$

$x^9$ এর সহগ $=(\frac{n}{9})2^{n-9}{\left(\frac{1}{5}\right)}^9$

শর্তমতে, $(\frac{n}{8})2^{n-8}{\left(\frac{1}{5}\right)}^8=(\frac{n}{9})2^{n-9}{\left(\frac{1}{5}\right)}^9$

বা, $\frac{n!}{8!(n-8)!}\frac{2^{n-8}}{2^{n-9}}=\frac{n!}{9!(n-9)!}\cdot \frac{{\left(\frac{1}{5}\right)}^9}{{\left(\frac{1}{5}\right)}^8}$

বা, $\frac{1}{8!(n-8)(n-9)!}{2}^{n-8-n+9}=\frac{1}{9\cdot 8!(n-9)!}{\left(\frac{1}{5}\right)}^{9-8}$

বা, $\frac{2}{n-8}=\frac{1}{9}\cdot \frac{1}{5}$

বা, $n-8=90$

বা, $n=90+8$

$\therefore n=98$

$\therefore (\frac{n}{2})=(\frac{98}{2})=\frac{98.97}{2.1}=4753$

নির্ণেয় মান : 4753.

দেওয়া আছে, $g\left(y\right)=\frac{2y^3}{1-y^3}=\frac{-2+2y^3+2}{1-y^3}=\frac{-2(1-y^3)+2}{1-y^3}$

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

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

ধরি, $\frac{2}{(1-y)(1+y+y^2)}\equiv \frac{A}{1-y}+\frac{By+C}{1+y+y^2}$ ......... (1)

(i) নং সমীকরণের উভয় পক্ষকে $(1-y)(1+y+y^2)$ দ্বারা গুণ করে পাই, $2\equiv A(1+y+y^2)+(\mathrm{By}+C)(1-y)$ ......... (2)

(2) নং সমীকরণে $y=1$ বসিয়ে পাই,

$2=A(1+1+1^2)+(B\times 1+C)(1-1)$

বা, $2=3A+0$

বা, $3A=2$

$\therefore A=\frac{2}{3}$

আবার, (2) নং সমীকরণ হতে পাই, $2=A+\mathrm{Ay}+{\mathrm{Ay}}^2+\mathrm{By}-\mathrm{By}{}^2+C-\mathrm{Cy}$

বা, $2=(A-B)y^2+(A+B-C)y+(A+C)$ ......... (3)

(3) নং সমীকরণ হতে $y^2$ ও $y$ এর সহগ সমৃকীত করে

$A-B=0$

বা, $A=B$

বা, $\frac{2}{3}=B$

$\therefore B=\frac{2}{3}$

আবার, $A+B-C=0$

বা, $\frac{2}{3}+\frac{2}{3}=C$

বা, $C=\frac{4}{3}$

(i) নং এ A, B ও C এর মান বসিয়ে পাই,

$\frac{2}{(1-y)(1+y+y^2)}\equiv \frac{\frac{2}{3}}{1-y}+\frac{\frac{2y}{3}+\frac{4}{3}}{1+y+y^2}\equiv \frac{2}{3(1-y)}+\frac{\frac{2y+4}{3}}{1+y+y^2}$

$\equiv \frac{2}{3(1-y)}+\frac{2y+4}{3(1+y+y^2)}$

$\therefore g\left(y\right)\equiv -2+\frac{2}{3(1-y)}+\frac{2y+4}{3(1+y+y^2)}$ যা নির্ণেয় আংশিক ভগ্নাংশ।