Answer:
A. 1, 4, 6, 4, 1
Step-by-step explanation:
Method 1
Binomial Series
[tex]\displaystyle (a+b)^n=a^n+\binom{n}{1}a^{n-1}b^1+\binom{n}{2}a^{n-2}b^2+...+\binom{n}{r}a^{n-r}b^r+...+b^n\\\\\\\textsf{Where }\displaystyle \binom{n}{r} \: = \:^{n}\text{C}_{r} = \frac{n!}{r!(n-r)!}[/tex]
Given expansion:
[tex](x + y)^4[/tex]
Therefore:
Substitute these values into the Binomial Series formula:
[tex]\implies \displaystyle (x+y)^4=x^4+\binom{4}{1}x^{4-1}y^1+\binom{4}{2}x^{4-2}y^2+\binom{4}{3}x^{4-3}y^3+\binom{4}{4}x^{4-4}y^4[/tex]
[tex]\implies \displaystyle (x+y)^4=x^4+\binom{4}{1}x^{3}y+\binom{4}{2}x^{2}y^2+\binom{4}{3}xy^3+\binom{4}{4}y^4[/tex]
[tex]\implies \displaystyle (x+y)^4=x^4+\frac{4!}{1!(4-1)!}x^{3}y+\frac{4!}{2!(4-2)!}x^{2}y^2+\frac{4!}{3!(4-3)!}xy^3+\frac{4!}{4!(4-4)!}y^4[/tex]
[tex]\implies \displaystyle (x+y)^4=x^4+\frac{4!}{1!3!}x^{3}y+\frac{4!}{2!2!}x^{2}y^2+\frac{4!}{3!1!}xy^3+\frac{4!}{4!0!}y^4[/tex]
[tex]\implies \displaystyle (x+y)^4=x^4+4x^{3}y+6x^{2}y^2+4xy^3+y^4[/tex]
Therefore, the coefficients corresponding to the terms of the expansion are:
Method 2
[tex]\boxed{\begin{minipage}{5cm} \underline{Binomial Theorem}\\\\$\displaystyle (a+b)^n=\sum^{n}_{k=0}\binom{n}{k} a^{n-k}b^{k}$\\\\\\where \displaystyle \binom{n}{k} = \frac{n!}{k!(n-k)!}\\\end{minipage}}[/tex]
Given expansion:
[tex](x + y)^4[/tex]
Therefore:
Substitute these values into the Binomial Theorem formula:
[tex]\displaystyle (x+y)^4=\sum^{4}_{k=0}\binom{4}{k} x^{4-k}y^{k}[/tex]
The coefficient of each term is given by:
[tex]\displaystyle \binom{4}{k} = \frac{4!}{k!(4-k)!}[/tex]
Therefore, to find the coefficients of the terms of the expansion, substitute k = 0 through 5 into the coefficient formula:
[tex]k=0 \implies \displaystyle \binom{4}{0} = \frac{4!}{0!(4-0)!}=1[/tex]
[tex]k=1 \implies \displaystyle \binom{4}{1} = \frac{4!}{1!(4-1)!}=4[/tex]
[tex]k=2 \implies \displaystyle \binom{4}{2} = \frac{4!}{2!(4-2)!}=6[/tex]
[tex]k=3 \implies \displaystyle \binom{4}{3} = \frac{4!}{3!(4-3)!}=4[/tex]
[tex]k=4 \implies \displaystyle \binom{4}{4} = \frac{4!}{4!(4-4)!}=1[/tex]
Therefore, the coefficients corresponding to the terms of the expansion are:
