Question

Using Binomial Theorem, evaluate $(102{)}^5$

Answer 1

$(102{)}^5=(100+2{)}^5$
${=}^5{C}_0\left(100{)}^5{+}^5{C}_1\right(100{)}^4\left(2\right){+}^5{C}_2\left(100{)}^3\right(2{)}^2{+}^5{C}_3\left(100{)}^2\right(2{)}^3{+}^5{C}_4\left(100\right)\left(2{)}^4{+}^5{C}_5\right(2{)}^5$
$=(100{)}^5+5(100{)}^4\left(2\right)+10\left(100{)}^3\right(2{)}^2+10\left(100{)}^2\right(2{)}^3+5\left(100\right)(2{)}^4+(2{)}^5$
$=10000000000+1000000000+40000000+800000+8000+32$

$=11040808032$