সৃজনশীলঃ ‘দৃশ্যকল্প-১ : f(p,q,r)=(p+q+r)(pq+qr+rp)দৃশ্যকল্প-২…’ ফৌজদারহাট ক্যাডেট কলেজ ২০২৬ – দ্বিপদী বিস্তৃতি, উচ্চতর গণিত (এসএসসি)

দেওয়া আছে, $f(p,q,r)=(p+q+r)(\mathrm{pq}+\mathrm{qr}+\mathrm{rp})f(p,q,r)=\mathrm{pqr}$ হলে,

$(p+q+r)(pq+qr+rp)=pqr$

বা, $(p+q+r)(pq+qr+rp)-pqr=0$

বা, $p(\mathrm{pq}+\mathrm{qr}+\mathrm{rp})+q(\mathrm{pq}+\mathrm{qr}+\mathrm{rp})+r(\mathrm{pq}+\mathrm{qr}+\mathrm{rp})-\mathrm{pqr}=0$

বা, $p^{2}q+pqr+r{p}^2+p{q}^2+q^{2}r+pqr+pqr+q{r}^2+r^{2}p-pqr=0$

বা, $p^{2}q+pqr+r{p}^2+p{q}^2+q^{2}r+pqr+q{r}^2+r^{2}p=0$

বা, $p^{2}q+p{q}^2+pqr+q^{2}r+r{p}^2+pqr+r^{2}p+q{r}^2=0$

বা, $pq(p+q)+qr(p+q)+rp(p+q)+r^2(p+q)=0$

বা, $(p+q)(pq+qr+rp+r^2)=0$

বা, $(p+q)\left\{q\right(p+r)+r(p+r\left)\right\}=0$

বা, $(p+q)(q+r)(p+r)=0$

হয়, $p+q=0$

$\therefore p=-q$

অথবা, $(q+r)=0$

বা, $q=-r$

অথবা, $p+r=0$ 

বা, $r=-p$

এখন, $p=-q$  হলে,

বামপক্ষ $=\frac{1}{(p+q+r{)}^{\Pi }}=\frac{1}{(-q+q+r{)}^{11}}=\frac{1}{r^{\Pi }}$

ডানপক্ষ $=\frac{1}{p^{\Pi }}+\frac{1}{q^{\Pi }}+\frac{1}{r^{\Pi }}=\frac{1}{(-q{)}^{11}}+\frac{1}{q^{\Pi }}+\frac{1}{r^{\Pi }}$

$=-\frac{1}{q^{\Pi }}+\frac{1}{q^{\Pi }}+\frac{1}{r^{\Pi }}=\frac{1}{r^{\Pi }}$

$\therefore \frac{1}{(p+q+r{)}^{\Pi }}=\frac{1}{p^{\Pi }}+\frac{1}{q^{\Pi }}+\frac{1}{r^{\Pi }.}$ (প্রমাণিত)

দেওয়া আছে, $g\left(x\right)=(2+\frac{x}{4})$

$\therefore \left\{g\right(x){\}}^7={(2+\frac{x}{4})}^7$

প্যাসকেলের ত্রিভুজের সাহায্যে $\left\{g\right(x){\}}^7={(2+\frac{x}{4})}^7$ এর বিস্তৃত করে পাই,

$r\therefore {(2+\frac{x}{4})}^7={1.2}^7+{7.2}^{7-1}\cdot {\left(\frac{x}{4}\right)}^1+{21.2}^{7-2}\cdot {\left(\frac{x}{4}\right)}^2$ 

$+{35.2}^{7-3}\cdot {\left(\frac{x}{4}\right)}^3+{35.2}^{7-4}\cdot {\left(\frac{x}{4}\right)}^4$ 

$+{21.2}^{7-5}\cdot {\left(\frac{x}{4}\right)}^5+{7.2}^{7-6}\cdot {\left(\frac{x}{4}\right)}^6+1\cdot {\left(\frac{x}{4}\right)}^7$

$=128+{7.2}^6\cdot \frac{x}{4}+{21.2}^5\cdot \frac{x^2}{16}+{35.2}^4\cdot \frac{x^3}{64}+{35.2}^3\cdot \frac{x^4}{256}$

$+{21.2}^2\cdot \frac{x^5}{1024}+{7.2}^1\cdot \frac{x^6}{4096}+\frac{x^7}{16384}=128+112x+42x^2+\frac{35}{4}{x}^3+\frac{35}{32}{x}^4+\frac{21}{256}{x}^5+\frac{7}{{2048}^x}{x}^6+\frac{x^7}{16384}$

এখন, ${(2+\frac{x}{4})}^{-7}=(1.01{)}^7$

বা, $2+\frac{x}{4}=1.01$

বা, $\frac{x}{4}=1.01-2=-0.99$

বা, $x=-0.99\times 4=-3.96$

$\therefore {\{2+\left(\frac{-3.96}{4}\right)\}}^7=128+112(-3.96)$

$+42(-3.96{)}^2+\frac{35}{4}(-3.96{)}^3+\frac{35}{32}(-3.96{)}^4$

$+\frac{21}{256}(-3.96{)}^5+\frac{7}{2048}(-3.96{)}^6+\frac{(-3.96{)}^7}{16384}$

$+\frac{7}{2048}(-3.96{)}^6+\frac{(-3.96{)}^7}{16384}$

বা, $(2-0.99{)}^7=128-443.52+658.6272-543.36744+268.9668828-79.88316419+13.18072209-0.9320653479$

বা, $(1.01{)}^7=1.072135352$

$\therefore (1.01{)}^7=1.072135$

নির্ণেয় মান 1.072135 (প্রায়)।