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

(i)$(102{)}^4=(100+2{)}^4$

${=}^4{C}_0\left(100{)}^4{+}^4{C}_1\right(100{)}^3\left(2\right){+}^4{C}_2\left(100{)}^2{2}^2{+}^4{C}_3\right(100\left)\right(2{)}^3{+}^4{C}_2(2{)}^4$

$=(100{)}^4+4.(100{)}^{3}2+6(100{)}^2{2}^2+4(100)2^3+2^3$

$100000000+8000000+240000+3200+16$

=108243216

$\left(ii\right)(1+0.1{)}^5$

${=}^5{C}_0{1}^5{+}^5{C}_1{1}^4\left(0.1\right){+}^2{C}_2{1}^3\left(0.1{)}^2{+}^5{C}_3{1}^2\right(0.1{)}^3{+}^5{C}_{4}1\left(0.1{)}^4{+}^5{C}_5\right(0.1{)}^5$

$=1+5\left(0.1\right)+10(0.1{)}^2+10(0.1{)}^3+5(0.1{)}^4+(0.1{)}^5$

=1+0.5+0.1+0.01+0.0005+0.00001

=1.61051