Question

Using Binomial theorem, evaluate $(99{)}^5$

Answer 1

$(99{)}^5=(100-1{)}^5$
${=}^5{C}_0\left(100{)}^5{-}^5{C}_1\right(100{)}^4\left(1\right){+}^5{C}_2\left(100{)}^3\right(1{)}^2{-}^5{C}_3\left(100{)}^2\right(1{)}^3{+}^5{C}_4\left(100\right)\left(1{)}^4{-}^5{C}_5\right(1{)}^5$
$=(100{)}^5-5(100{)}^4+10(100{)}^3-10(100{)}^2+5\left(100\right)-1$
$=10000000000-500000000+10000000-100000+500-1$
$=10010000500-500100001$

$=9509900499$