Question

use pascals triangle to complete the expansion of $(y - z)^4$.
$y^4 + \square y^{\square} z + 6y^2 z^2 - 4yz^3 + z^4$

Explanation:

Step1: Recall Pascal's Triangle for power 4

Pascal's Triangle for the 4th power (since we have \((y - z)^4\)) has coefficients 1, 4, 6, 4, 1. The general binomial expansion is \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\). For \((y - z)^4=(y+(-z))^4\), the expansion is \(y^{4}+4y^{3}(-z)+6y^{2}(-z)^{2}+4y(-z)^{3}+(-z)^{4}\).

Step2: Simplify the terms

Simplify each term:

• The second term: \(4y^{3}(-z)=- 4y^{3}z\)? Wait, no, wait. Wait, the given expansion has a sign error? Wait, no, let's re - check. Wait, the given expansion is \(y^{4}+\square y^{\square}z + 6y^{2}z^{2}-4yz^{3}+z^{4}\). Wait, actually, when we expand \((y - z)^4\), using the binomial theorem \((a - b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}(-b)^{k}\). So for \(n = 4\), the coefficients are 1, 4, 6, 4, 1. So the expansion is:

\(y^{4}+4y^{3}(-z)+6y^{2}z^{2}+4y(-z)^{3}+z^{4}\)
\(=y^{4}-4y^{3}z + 6y^{2}z^{2}-4yz^{3}+z^{4}\). But the given expansion has a plus sign before the second term. Wait, maybe there is a typo, but looking at the given expansion: \(y^{4}+\square y^{\square}z + 6y^{2}z^{2}-4yz^{3}+z^{4}\). Wait, maybe the original problem has a sign mistake, but let's look at the coefficients. The coefficient of \(y^{3}z\) term in the binomial expansion of \((y - z)^4\) (when we consider the sign) - but the given expansion has a plus for the second term and a minus for the fourth term. Wait, actually, the binomial coefficients for \((a + b)^4\) are 1, 4, 6, 4, 1. If we have \((y+(-z))^4\), the coefficients of the terms are 1, 4, 6, 4, 1, but with alternating signs. But the given expansion has \(+ \square y^{\square}z\), \(+6y^{2}z^{2}\), \(-4yz^{3}\), \(+z^{4}\). Wait, the exponent of \(y\) in the second term: in the binomial expansion, for the term with \(y^{n - k}z^{k}\), when \(k = 1\), \(n=4\), so \(n - k=3\), so the exponent of \(y\) is 3. And the coefficient: from Pascal's triangle, the second coefficient (for \(k = 1\)) is 4. But since we have \((y - z)^4\), the term is \(4y^{3}(-z)=-4y^{3}z\), but the given expansion has a plus sign. Wait, maybe the problem has a typo, and it's supposed to be \((y + z)^4\)? No, because the last term is \(z^{4}\) and the fourth term is \(-4yz^{3}\). Wait, no, let's re - evaluate. Wait, the key is the coefficients from Pascal's triangle for \(n = 4\) are 1, 4, 6, 4, 1. So the second term (when \(k = 1\)) has a coefficient of 4, and the exponent of \(y\) is \(4-1 = 3\). Even though there is a sign issue in the given expansion (maybe a mistake in the sign of the second term), the coefficient and the exponent: the coefficient is 4 (but with a negative sign? Wait, no, the given expansion has \(+ \square y^{\square}z\) and then \(-4yz^{3}\). Wait, maybe the original problem is \((y + z)^4\) but with a sign error in the fourth term? No, \((y + z)^4=y^{4}+4y^{3}z + 6y^{2}z^{2}+4yz^{3}+z^{4}\). But the given has \(-4yz^{3}\). So perhaps the problem is \((y - z)^4\) and there is a sign error in the second term's sign in the problem statement. But according to the binomial theorem, for \((y - z)^4\), the expansion is \(y^{4}-4y^{3}z + 6y^{2}z^{2}-4yz^{3}+z^{4}\). But the problem's given expansion has \(+ \square y^{\square}z\), so maybe it's a typo and the sign of the second term is wrong, but the coefficient and exponent: the coefficient is 4 (absolute value) and the exponent of \(y\) is 3.

Answer:

The coefficient is \(- 4\) and the exponent of \(y\) is \(3\). Wait, but the given expansion has a plus sign. Wait, maybe the problem is written incorrectly. But according to the binomial expansion of \((y - z)^4\), the second term is \(-4y^{3}z\), so the box for the coefficient is \(-4\) and the exponent of \(y\) is \(3\).