Question

use pascals triangle to complete the expansion of (q - r)^6. q^6 - 6q^5r + 15q^4r^2 - 20q^3r^3 + \square q^2r^\square - 6qr^5 + r^6

Explanation:

Step1: Recall Pascal's Triangle for power 6

Pascal's Triangle row for \(n = 6\) is \(1, 6, 15, 20, 15, 6, 1\). Also, the sign alternates starting with positive for the first term, then negative, positive, etc. For the term with \(q^{2}r^{\square}\), the coefficient comes from Pascal's Triangle. The position of the term: in the expansion of \((a + b)^n\), the \(k\)-th term (starting from \(k = 0\)) has coefficient from Pascal's Triangle. For \((q - r)^6=\sum_{k = 0}^{6}\binom{6}{k}q^{6 - k}(-r)^{k}\). When \(6 - k=2\), then \(k = 4\). The binomial coefficient \(\binom{6}{4}=\frac{6!}{4!(6 - 4)!}=\frac{6\times5}{2\times1}=15\). And the power of \(r\) is \(k = 4\), and since \((-r)^{4}=r^{4}\), the sign is positive.

Step2: Determine the coefficient and the exponent of \(r\)

From Pascal's Triangle, the coefficient for the term where the power of \(q\) is \(2\) (i.e., \(6 - k = 2\), so \(k = 4\)) is \(15\), and the exponent of \(r\) is \(4\) (because \(k = 4\) in \((-r)^{k}\), and \((-r)^{4}=r^{4}\)). Also, the sign: since \(k = 4\) is even, the sign is positive. So the term is \(15q^{2}r^{4}\).

Answer:

The coefficient is \(15\) and the exponent of \(r\) is \(4\), so the boxed coefficient is \(15\) and the boxed exponent of \(r\) is \(4\). So the term is \(15q^{2}r^{4}\), so the first box (coefficient) is \(15\) and the second box (exponent of \(r\)) is \(4\).