Solution: 2017 Fall Midterm - 9

Author: Michiel Smid

Question

What is the coefficient of $x^{20}y^{80}$ in the expansion of $(7x-13y)^{100}$?

(a)

${100 \choose 80} \cdot 7^{20} \cdot 13^{80}$

(b)

$- {100 \choose 80} \cdot 7^{20} \cdot 13^{80}$

(c)

$- {100 \choose 20} \cdot 7^{80} \cdot 13^{20}$

(d)

${100 \choose 20} \cdot 7^{80} \cdot 13^{20}$

$ = \sum^{100}_{k=0} \binom{100}{k} {(7x)}^{n-k} {(-13y)}^{k} $

We only consider $k=80$, as it results in $y^{80}$.

$ = \binom{100}{80} \cdot {(7x)}^{100-80} \cdot {(-13y)}^{80} $

$ = \binom{100}{80} \cdot {(7)}^{20} \cdot x^{20} \cdot {(-13)}^{80} \cdot y^{80} $

$ = \binom{100}{80} \cdot 7^{20} \cdot 13^{80} \cdot x^{20} \cdot y^{80} $

Thus, the coefficient is $ \binom{100}{80} \cdot 7^{20} \cdot 13^{80} $