Solution: 2015 Fall Final - 4
Author: Michiel SmidQuestion
What is the coefficient of $x^{15}y^{5}$ in the expansion of $(-3x + 5y)^{20}$?
(a)
${20 \choose 5} \cdot 3^{15} \cdot 5^5$
(b)
${20 \choose 5} \cdot 5^{15} \cdot 3^5$
(c)
$- {20 \choose 5} \cdot 3^{15} \cdot 5^5$
(d)
$- {20 \choose 5} \cdot 5^{15} \cdot 3^5$
Solution
$ = \sum^{20}_{k=0} \binom{20}{k} {(-3x)}^{n-k} {(5y)}^{k} $
We only consider $k=5$, as it results in $y^{5}$.
$ = \binom{20}{5} \cdot {(-3x)}^{20-5} \cdot {(5y)}^{5} $
$ = \binom{20}{5} \cdot {(-3)}^{15} \cdot {5}^{5} \cdot x^{15} \cdot y^5 $
$ = - \binom{20}{5} \cdot {3}^{15} \cdot {5}^{5} \cdot x^{15} \cdot y^5 $
Thus, the coefficient of $x^{15}y^{5}$ in the expansion of ${(-3x + 5y)}^{20}$ is $ - \binom{20}{5} \cdot {3}^{15} \cdot {5}^{5} $