$a_1,a_2,a_3,a_4$ যদি ${\left(1+x\right)}^n$ এর বিকৃতির চারটি ক্রমিক পদের সহগ হয়ে থাকে তাহলে প্রমাণ কর যে, $\frac{a_1}{a_1+a_2}+\frac{a_2}{a_3+a_4}=\frac{2a_2}{a_2+a_3}$

প্রদত্ত রাশি:  ${\left(2-\frac{1}{3x}\right)}^5$

দ্বিপদী উপপাদ্যের সাহায্যে বিস্তৃতি করে পাই,

${\left(2-\frac{1}{3x}\right)}^5=2^5{+}^5{C}_1{2}^{5-1}{\left(-\frac{1}{3x}\right)}^1{+}^5{C}_2{2}^{5-2}{\left(-\frac{1}{3x}\right)}^2{+}^5{C}_3{2}^{5-3}{\left(-\frac{1}{3x}\right)}^3{+}^5{C}_4{2}^{5-4}{\left(-\frac{1}{3x}\right)}^4{+}^5{C}_5{2}^{5-5}{\left(-\frac{1}{3x}\right)}^5$

$=32+5\times 2^4\times \frac{-1}{3x}+10\times 2^3\times \frac{1}{9x^2}+10\times 2^2\times \frac{-1}{27x^3}+5\times 2^1\times \frac{1}{81x^4}+1\times 2^0\times \frac{-1}{243\times x^5}$

$=32-\frac{80}{3x}+\frac{80}{9x^2}-\frac{80}{27x^3}+\frac{10}{81x^4}-\frac{-1}{243x^5}$

নির্ণেয় বিস্তৃতি:  $32-\frac{80}{3x}+\frac{80}{9x^2}-\frac{80}{27x^3}+\frac{10}{81x^4}-\frac{-1}{243x^5}$

প্রদত্ত রাশি:  ${\left(y-\frac{1}{y^2}\right)}^5$

দ্বিপদী উপপাদ্যের সাহায্যে বিস্তৃতি করে পাই,

${\left(y-\frac{1}{y^2}\right)}^5=y^5 +{ }^5{C}_1{y}^{5-1}{\left(-\frac{1}{y^2}\right)}^1{+}^5{C}_2{y}^{5-2}{\left(-\frac{1}{y^2}\right)}^2{+}^5{C}_3{y}^{5-3}{\left(-\frac{1}{y^2}\right)}^3{+}^5{C}_4{y}^{5-4}{\left(-\frac{1}{y^2}\right)}^4{+}^5{C}_5{y}^{5-5}{\left(-\frac{1}{y^2}\right)}^5$

$=y^5 + 5y^4. \frac{1}{y^2}+{\frac{5.4}{1.2}}^2{y}^3.\frac{1}{y^4}-\frac{5.43}{1.2.3}{y}^2.\frac{1}{y^4}-\frac{5.4.3}{1.2.3}{y}^2.\frac{1}{y^6}-\frac{5.4.3.2}{1.2.3.4}y.\frac{1}{y^x}-\frac{1}{y^{10}}$

$=y^5-5y^2+ \frac{10}{y}-{\frac{10}{y^4}}^2+.\frac{5}{y^7}-\frac{1}{y^{10}}$

নির্ণেয় বিস্তৃতি  $=y^5-5y^2+ \frac{10}{y}-{\frac{10}{y^4}}^2+.\frac{5}{y^7}-\frac{1}{y^{10}}$

দ্বিপদী উপপাদ্য অনুসারে পাই,

${\left(x+\frac{1}{x^2}\right)}^4=x^6+\left(\frac{6}{1}\right)x^5\left(\frac{1}{x^2}\right)+\left(\frac{6}{2}\right)x^4{\left(\frac{1}{x^2}\right)}^2+\left(\frac{6}{3}\right)x^3{\left(\frac{1}{x^2}\right)}^3+......$

$=x^6+6x^3+\frac{6.5}{1.2}{x}^4 \frac{1}{x^4}+\frac{6.5.4}{1.2.3}{x}^3\frac{1}{x^6}+......$

$=x^6+6x^3+15+20\frac{1}{x^3}+......$

নির্ণেয় বিস্তৃতি : $x^6+6x^3+15+20\frac{1}{x^3}+......$

দ্বিপদী উপপাদ্য ব্যবহার করে,

${\left(2x^2-\frac{1}{x^2} \right)}^8={\left(2x^2\right)}^8+{ }^8{C}_1{\left(2x^2\right)}^7 \left(-\frac{1}{x^2}\right){+}^8{C}_2{\left(2x^2\right)}^6 {\left(-\frac{1}{x^2}\right)}^2{+}^8{C}_3{\left(2x^2\right)}^5 {\left(-\frac{1}{x^2}\right)}^3+.....$

$=2^8.x^{16}-8.2^7.x^{14}.\frac{1}{x^2}+\frac{8.7}{1.2}.2^6.x^{12}.\frac{1}{x^4}-\frac{8.7.6}{1.2.3}.2^5.x^{10}.\frac{1}{x^6}+.....$

$=256x^{16}-1024x^{12}+1792x^8-1792x^4+.....$

নির্ণেয় বিস্তৃতি: $256x^{16}-1024x^{12}+1792x^8-1792x^4+.....$

k = 1 হলে, বীজগাণিতিক রাশিটি ${\left(1-\frac{x}{3}\right)}^7$

দ্বিপদী উপপাদ্য অনুসারে পাই,

${\left(1-\frac{x}{3}\right)}^7{=}^7{C}_{0 }{\left(-\frac{x}{3}\right)}^0{+}^7{C}_{1 }{\left(-\frac{x}{3}\right)}^1{+}^7{C}_{2 }{\left(-\frac{x}{3}\right)}^2{+}^7{C}_{3 }{\left(-\frac{x}{3}\right)}^3+...$

$=1-7.\frac{x}{3}+\frac{7.6}{1.2}.\frac{x^2}{9}-\frac{7.6.5}{1.2.3}.\frac{x^3}{27}+...$

$=1-\frac{7x}{3}+\frac{7x^2}{3}-\frac{35x^3}{27}+......$

নির্ণেয় বিস্তৃতি: $1-\frac{7x}{3}+\frac{7x^2}{3}-\frac{35x^3}{27}+......$

দ্বিপদী উপপাদ্যের সাহায্যে বিস্তৃতি করে পাই,

${\left(P+\frac{x}{2}\right)}^6=P^6+{ }^6{C}_1{P}^{6-1} {\left(\frac{x}{2}\right)}^1+{ }^6{C}_2{P}^{6-2} {\left(\frac{x}{2}\right)}^2+......$