Question

Using binomial theorem, find the value of
$(i)\,\,{{(102)}^{4}}$
$(ii)\,\,{{(1.1)}^{5}}$

Answer 1

We are going to use the binomial theorem to expand the given values. After expanding the terms, we will get the required answer.

Formula used:
Formula is used for the binomial theorem
${{(x+a)}^{n}}{{=}^{n}}{{C}_{0}}{{x}^{n}}{{+}^{n}}{{C}_{1}}{{(x)}^{n1}}a{{+}^{n}}{{C}_{2}}{{(x)}^{n2}}{{a}^{2}}+.......{{+}^{n}}{{C}_{n}}{{a}^{n}}$

Complete step by step answer:
Formula is used for the binomial theorem
${{(x+a)}^{n}}{{=}^{n}}{{C}_{0}}{{x}^{n}}{{+}^{n}}{{C}_{1}}{{(x)}^{n1}}a{{+}^{n}}{{C}_{2}}{{(x)}^{n2}}{{a}^{2}}+.......{{+}^{n}}{{C}_{n}}{{a}^{n}}$
{{(100+2)}^{4}}{{=}^{4}}{{C}_{0}}{{(100)}^{4}}{{+}^{4}}{{C}_{1}}{{(100)}^{41}}(2){{+}^{4}}{{C}_{2}}$$$${{(100)}^{42}}{{(2)}^{2}}{{+}^{4}}{{C}_{3}}{{(100)}^{43}}{{(2)}^{3}}+$^{4}{{C}_{4}}{{(2)}^{4}}$
Rewrite the expression after simplification

4 \,}} \right. }{\left| \\!{\underline {\, 0 \,}} \right. \times \left| \\!{\underline {\, 4 \,}} \right. - 0}{{(100)}^{4}}+\dfrac{\left| \\!{\underline {\, 4 \,}} \right. }{\left| \\!{\underline {\, 1 \,}} \right. \times \left| \\!{\underline {\, 4 \,}} \right. - 1}\times {{(100)}^{3}}2$$+$$\dfrac{\left| \\!{\underline {\, 4 \,}} \right. }{\left| \\!{\underline {\, 2 \,}} \right. \times \left| \\!{\underline {\, 4 \,}} \right. - 2}{{(100)}^{2}}\times 4+\dfrac{\left| \\!{\underline {\, 4 \,}} \right. }{\left| \\!{\underline {\, 3 \,}} \right. \times \left| \\!{\underline {\, 4 \,}} \right. - 3}(100)\times 8+16$$ Simplify the expression $$\Rightarrow \,\,\,{{(100)}^{4}}+\dfrac{4\times \left| \\!{\underline {\, 3 \,}} \right. }{\left| \\!{\underline {\, 1 \,}} \right. \times \left| \\!{\underline {\, 3 \,}} \right. }\,\,\,100\times 8+16$$ Rewrite the equation after simplification $$=\,\,\,100000000+8000000+240000+3200+16$$ $$=108243216$$ $$(ii)\,\,\,{{(1+0.1)}^{5}}$$ Use the formula of the binomial theorem $${{(x+a)}^{n}}=\,{{\,}^{n}}{{C}_{0}}{{x}^{n}}a{{+}^{n}}{{C}_{1}}{{x}^{n1}}a{{+}^{n}}{{C}_{2}}{{x}^{n2}}{{a}^{2}}+.......{{+}^{n}}{{C}_{n}}{{a}^{n}}$$ $${{(1+0.1)}^{5}}{{=}^{5}}{{C}_{0}}{{(1)}^{5}}{{+}^{5}}{{C}_{1}}{{(1)}^{51}}(0.1){{+}^{5}}{{C}_{2}}{{(1)}^{51}}{{(0.1)}^{2}}$$+$$^{5}{{C}_{3}}{{(1)}^{53}}{{(0.1)}^{3}}{{+}^{5}}{{C}_{4}}{{(1)}^{54}}{{(0.1)}^{4}}{{(100)}^{2}}\times 4{{+}^{5}}{{C}_{5}}{{(0.1)}^{5}}$$ Simplify the expression $$\Rightarrow \,\,\,1+5\times (0.1)+\dfrac{\left| \\!{\underline {\, 5 \,}} \right. }{\left| \\!{\underline {\, 2 \,}} \right. \,\times \left| \\!{\underline {\, 3 \,}} \right. }{{(0.1)}^{2}}+\dfrac{\left| \\!{\underline {\, 5 \,}} \right. }{\left| \\!{\underline {\, 3 \,}} \right. \times \left| \\!{\underline {\, 2 \,}} \right. }\times {{(0.1)}^{3}}$$+ $$\dfrac{\left| \\!{\underline {\, 5 \,}} \right. }{\left| \\!{\underline {\, 4 \,}} \right. \times \left| \\!{\underline {\, 1 \,}} \right. }\times 1\times {{(0.1)}^{4}}+{{(0.1)}^{5}}$$ Simplify the expression $$\Rightarrow \,\,1+5\times (0.1)+\dfrac{\left| \\!{\underline {\, 3 \,}} \right. \times 4\times 5}{1\times 2\times 3}{{(0.1)}^{2}}+\dfrac{\left| \\!{\underline {\, 3 \,}} \right. \times 4\times 5}{\left| \\!{\underline {\, 3 \,}} \right. \times 2\times 1}\times (0.1)$$+$$\dfrac{\left| \\!{\underline {\, 5 \,}} \right. }{\left| \\!{\underline {\, 4 \,}} \right. \times \left| \\!{\underline {\, 1 \,}} \right. }\times {{(0.1)}^{4}}+{{(0.1)}^{5}}$$ Rewrite the expression after simplification $$\Rightarrow \,\,1+0.5+10\times {{(0.1)}^{2}}+10\times {{(0.1)}^{3}}+5{{(0.1)}^{4}}+{{(0.1)}^{5}}$$ Use the concept of the addition $$\Rightarrow \,\,1+0.5+0.1+0.01+0.0005+0.00001$$ $$\Rightarrow \,\,1.61051$$ **Note:** $$(i)$$These types of problems are always solved by the binomial theorem. $$(ii)$$When the concept of the binomial theorem is used, then we always use the factorial method.