Question

use pascals triangle to complete the expansion of $(p + q)^6$.
$p^6 + 6p^5q + \square p^4q^2 + 20p^3q^3 + \square p^2q^4 + 6pq^5 + q^6$

Explanation:

Step1: Recall Pascal's Triangle for \(n = 6\)

Pascal's Triangle rows correspond to binomial coefficients \(\binom{n}{k}\) for \((a + b)^n\). For \(n = 6\), the row of Pascal's Triangle is \(1, 6, 15, 20, 15, 6, 1\). These are the coefficients for \((p + q)^6=\sum_{k = 0}^{6}\binom{6}{k}p^{6 - k}q^{k}\).

Step2: Find coefficient for \(p^{4}q^{2}\)

For the term \(p^{4}q^{2}\), \(k = 2\) (since \(q\) has exponent \(2\)). The binomial coefficient \(\binom{6}{2}=\frac{6!}{2!(6 - 2)!}=\frac{6\times5}{2\times1}=15\). So the coefficient for \(p^{4}q^{2}\) is \(15\).

Step3: Find coefficient for \(p^{2}q^{4}\)

For the term \(p^{2}q^{4}\), \(k = 4\) (since \(q\) has exponent \(4\)). The binomial coefficient \(\binom{6}{4}=\frac{6!}{4!(6 - 4)!}=\frac{6\times5}{2\times1}=15\) (or note that \(\binom{n}{k}=\binom{n}{n - k}\), so \(\binom{6}{4}=\binom{6}{2}=15\)). So the coefficient for \(p^{2}q^{4}\) is \(15\).

Answer:

The first blank (coefficient of \(p^{4}q^{2}\)) is \(15\) and the second blank (coefficient of \(p^{2}q^{4}\)) is \(15\). So the filled expansion is \(p^{6}+6p^{5}q + 15p^{4}q^{2}+20p^{3}q^{3}+15p^{2}q^{4}+6pq^{5}+q^{6}\).