Respuesta :
to solve the questions we proceed as follows:
(2x+y)^5
=(2x+y)^2(2x+y)^2(2x+y)
=(4x^2+4xy+y^2)(4x^2+4xy+y^2)(2x+y)
=32x^5+80x^4y+80x^3y^2+40x^2y^3+10xy^4+y^5
the coefficient of xy^4 is 10
the answer is 10
(2x+y)^5
=(2x+y)^2(2x+y)^2(2x+y)
=(4x^2+4xy+y^2)(4x^2+4xy+y^2)(2x+y)
=32x^5+80x^4y+80x^3y^2+40x^2y^3+10xy^4+y^5
the coefficient of xy^4 is 10
the answer is 10
Answer:
The coefficient of [tex]xy^4[/tex] in the expension of [tex](2x+y)^5[/tex] is 10.
Step-by-step explanation:
Given [tex](2x+y)^5[/tex] , we have to find the coefficient of [tex]xy^4[/tex] in the expension of [tex](2x+y)^5[/tex].
[tex]\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i[/tex]
Here, a = 2x and b = y ,
Substitute above, we get,
[tex]=\sum _{i=0}^5\binom{5}{i}\left(2x\right)^{\left(5-i\right)}y^i[/tex]
Using [tex]\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}[/tex]
[tex]\quad \:i=0\quad :\quad \frac{5!}{0!\left(5-0\right)!}\left(2x\right)^5y^0[/tex]
[tex]\quad \:i=1\quad :\quad \frac{5!}{1!\left(5-1\right)!}\left(2x\right)^4y^1\\\\\quad \:i=2\quad :\quad \frac{5!}{2!\left(5-2\right)!}\left(2x\right)^3y^2\\\\\ \quad \:i=3\quad :\quad \frac{5!}{3!\left(5-3\right)!}\left(2x\right)^2y^3\\\\\quad \:i=4\quad :\quad \frac{5!}{4!\left(5-4\right)!}\left(2x\right)^1y^4\\\\\quad \:i=5\quad :\quad \frac{5!}{5!\left(5-5\right)!}\left(2x\right)^0y^5\\\\[/tex]
Adding all terms, we get,
[tex](2x+y)^5=\quad \frac{5!}{0!\left(5-0\right)!}\left(2x\right)^5y^0+\quad \frac{5!}{1!\left(5-1\right)!}\left(2x\right)^4y^1+\quad \frac{5!}{2!\left(5-2\right)!}\left(2x\right)^3y^2+\quad \frac{5!}{3!\left(5-3\right)!}\left(2x\right)^2y^3+\quad \frac{5!}{4!\left(5-4\right)!}\left(2x\right)^1y^4+\quad \frac{5!}{5!\left(5-5\right)!}\left(2x\right)^0y^5[/tex]
On evaluating , we get,
[tex](2x+y)^5=32x^5+80x^4y+80x^3y^2+40x^2y^3+10xy^4+y^5[/tex]
Thus, the coefficient of [tex]xy^4[/tex] in the expension of [tex](2x+y)^5[/tex] is 10.
