Question

1. expand the binomial $(x + y)^4$ by applying pascals triangle.

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
$x^4y + 4x^3y^2 + 6x^2y^3 + 4xy^4 + y^5$
$x^4y^4 + 4x^3y^3 + 6x^2y^2 + 4xy + y^4$
$x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$
$x^4y + 3x^3y^2 + 3x^2y^3 + xy^4$

Explanation:

Step1: Identify the coefficients from Pascal's triangle

For \((a + b)^n\), the coefficients of the expansion can be found from the \((n + 1)\) - th row of Pascal's triangle. For \(n=4\), the row is \(1,4,6,4,1\).

Step2: Apply the binomial expansion formula \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\)

Here \(a=x\), \(b = y\), and \(n = 4\).
When \(k = 0\): \(\binom{4}{0}x^{4-0}y^{0}=1\times x^{4}\times1=x^{4}\)
When \(k = 1\): \(\binom{4}{1}x^{4 - 1}y^{1}=4\times x^{3}y\)
When \(k = 2\): \(\binom{4}{2}x^{4-2}y^{2}=6\times x^{2}y^{2}\)
When \(k = 3\): \(\binom{4}{3}x^{4-3}y^{3}=4\times xy^{3}\)
When \(k = 4\): \(\binom{4}{4}x^{4-4}y^{4}=1\times x^{0}y^{4}=y^{4}\)

Step3: Combine the terms

\((x + y)^4=x^{4}+4x^{3}y + 6x^{2}y^{2}+4xy^{3}+y^{4}\)

Answer:

\(x^{4}+4x^{3}y + 6x^{2}y^{2}+4xy^{3}+y^{4}\) (corresponds to the third option \(x^{4}+4x^{3}y+6x^{2}y^{2}+4xy^{3}+y^{4}\))