Solution: 2014 Winter Midterm - 7
Author: Michiel SmidQuestion
What is the coefficient of $x^{9}y^{16}$ in the expansion of $(7x + 21y)^{25}$?
(a)
${25 \choose 16} 7^{16}21^{9}$
(b)
${16 \choose 25} 7^{9}21^{16}$
(c)
${25 \choose 16} 7^{25}3^{16}$
(d)
none of the above
Solution
$=\sum^{25}_{k=0} \binom{25}{k}{(7x)}^k {(21y)}^{25-k}$
$=\binom{25}{16}{(7x)}^{9} {(21y)}^{16}$
$=\binom{25}{16}7^{9} x^{9} {21}^{16} y^{16}$
$=\binom{25}{16}7^{9} x^{9} 3^{16} 7^{16} y^{16}$
$=\binom{25}{16}7^{25} 3^{16} x^{9} y^{16}$
Thus, the coefficient is $\binom{25}{16}7^{25} 3^{16}$