Question

Let the numerically greatest term in the expansion of ${(3-2x)}^9$ When $x=1$ be $k$.Find last digit of $k$ ?

Answer 1

$(3-2x{)}^9$
$=3^9(1-\frac{2x}{3}{)}^9$
Now let $x=1$
We get
$r=\frac{(1+9).\frac{2}{3}}{1+\frac{2}{3}}$
$=\frac{20}{5}=4$
Hence coefficient of
$T_5$
$=T_{4+1}$
$={}^9{C}_4{.3}^5{.2}^4$
$=126\left(243\right).16$
$=489888$
Hence last digit is 8.