Question

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

Answer 1

Hint: 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
$$\Rightarrow {\displaystyle \frac{|\underset{―}{4}}{|\underset{―}{0}\times |\underset{―}{4}-0}}{(100)}^4+{\displaystyle \frac{|\underset{―}{4}}{|\underset{―}{1}\times |\underset{―}{4}-1}}\times {(100)}^{3}2$$+$${\displaystyle \frac{|\underset{―}{4}}{|\underset{―}{2}\times |\underset{―}{4}-2}}{(100)}^2\times 4+{\displaystyle \frac{|\underset{―}{4}}{|\underset{―}{3}\times |\underset{―}{4}-3}}(100)\times 8+16$$
Simplify the expression
$$\Rightarrow {(100)}^4+{\displaystyle \frac{4\times |\underset{―}{3}}{|\underset{―}{1}\times |\underset{―}{3}}}100\times 8+16$$
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
$$\Rightarrow 1+5\times (0.1)+{\displaystyle \frac{|\underset{―}{5}}{|\underset{―}{2}\times |\underset{―}{3}}}{(0.1)}^2+{\displaystyle \frac{|\underset{―}{5}}{|\underset{―}{3}\times |\underset{―}{2}}}\times {(0.1)}^3$$+ $${\displaystyle \frac{|\underset{―}{5}}{|\underset{―}{4}\times |\underset{―}{1}}}\times 1\times {(0.1)}^4+{(0.1)}^5$$
Simplify the expression
$$\Rightarrow 1+5\times (0.1)+{\displaystyle \frac{|\underset{―}{3}\times 4\times 5}{1\times 2\times 3}}{(0.1)}^2+{\displaystyle \frac{|\underset{―}{3}\times 4\times 5}{|\underset{―}{3}\times 2\times 1}}\times (0.1)$$+$${\displaystyle \frac{|\underset{―}{5}}{|\underset{―}{4}\times |\underset{―}{1}}}\times {(0.1)}^4+{(0.1)}^5$$
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.