Question

Question 7 of 25

The coefficients corresponding to k = 0, 1, 2, ..., 4 in the expansion of (x + y)²
are
A. 1, 4, 6, 4, 1
B. 0, 1, 4, 6, 4, 1,0
C. 0, 4, 6, 4, 0
D. 1,4,6,6,4,1

Answer 1

A. 1, 4, 6, 4, 1

Explanation:

Method 1

Binomial Series

`\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) = (n!)/(r!(n-r)!)`

Given expansion:

`(x + y)^4`

Therefore:

• a = x
• b = y
• n = 4

Substitute these values into the Binomial Series formula:

`\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`

`\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`

`\implies \displaystyle (x+y)^4=x^4+(4!)/(1!(4-1)!)x^(3)y+(4!)/(2!(4-2)!)x^(2)y^2+(4!)/(3!(4-3)!)xy^3+(4!)/(4!(4-4)!)y^4`

`\implies \displaystyle (x+y)^4=x^4+(4!)/(1!3!)x^(3)y+(4!)/(2!2!)x^(2)y^2+(4!)/(3!1!)xy^3+(4!)/(4!0!)y^4`

`\implies \displaystyle (x+y)^4=x^4+4x^(3)y+6x^(2)y^2+4xy^3+y^4`

Therefore, the coefficients corresponding to the terms of the expansion are:

• 1, 4, 6, 4, 1

Method 2

`\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} = (n!)/(k!(n-k)!)\\\end{minipage}}`

Given expansion:

`(x + y)^4`

Therefore:

• a = x
• b = y
• n = 4

Substitute these values into the Binomial Theorem formula:

`\displaystyle (x+y)^4=\sum^(4)_(k=0)\binom{4}{k} x^(4-k)y^(k)`

The coefficient of each term is given by:

`\displaystyle \binom{4}{k} = (4!)/(k!(4-k)!)`

Therefore, to find the coefficients of the terms of the expansion, substitute k = 0 through 5 into the coefficient formula:

`k=0 \implies \displaystyle \binom{4}{0} = (4!)/(0!(4-0)!)=1`

`k=1 \implies \displaystyle \binom{4}{1} = (4!)/(1!(4-1)!)=4`

`k=2 \implies \displaystyle \binom{4}{2} = (4!)/(2!(4-2)!)=6`

`k=3 \implies \displaystyle \binom{4}{3} = (4!)/(3!(4-3)!)=4`

`k=4 \implies \displaystyle \binom{4}{4} = (4!)/(4!(4-4)!)=1`

Therefore, the coefficients corresponding to the terms of the expansion are:

• 1, 4, 6, 4, 1