Answer :
Answer:
Remember, the expansion of [tex](x+y)^n[/tex] is [tex](x+y)^n=\sum_{k=0}^n \binom{n}{k}x^{n-k}y^k[/tex], where [tex]\binom{n}{k}=\frac{n!}{(n-k)!k!}[/tex].
Then,
[tex](5p+2q)^6=\sum_{k=0}^6\binom{6}{k}(5p)^{6-k}(2q)^k=\sum_{k=0}^6\binom{6}{k}5^{6-k}2^k p^{6-k}q^k[/tex]
Then, the coefficient of the term [tex]p^{6-k}q^k[/tex] is [tex]\binom{6}{k}5^{6-k}2^k[/tex]
a) since 6-k=2, then k=4. So the coefficient of [tex]p^2q^4[/tex] is
[tex]\binom{6}{4}5^{6-4}2^4=15*5^2*2^4=15*25*16=6000[/tex]
b) since 6-k=5, then k=1. So, the coefficient of [tex]p^5q[/tex] is
[tex]\binom{6}{1}5^{6-1}2^1=6*5^5*2=37500[/tex]
c) since 6-k=3, then k=3. So, the coefficient of [tex]p^3q^3[/tex] is
[tex]\binom{6}{3}5^{6-3}2^3=20*5^3*8=20000[/tex]