Option C:
The coefficient of [tex]x^{2} y^{3}[/tex] is 40.
Solution:
Given expression:
[tex](2 x+y)^{5}[/tex]
Using binomial theorem:
[tex](a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}[/tex]
Here [tex]a=2 x, b=y[/tex]
Substitute in the binomial formula, we get
[tex](2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}[/tex]
Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.
[tex]$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}[/tex]
[tex]$+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}[/tex]
Let us solve the term one by one.
[tex]$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}[/tex]
[tex]$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y[/tex]
[tex]$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}[/tex]
[tex]$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}[/tex]
[tex]$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}[/tex]
[tex]$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}[/tex]
Substitute these into the above expansion.
[tex](2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}[/tex]
The coefficient of [tex]x^{2} y^{3}[/tex] is 40.
Option C is the correct answer.
