Question

use pascals triangle to complete the expansion of $(t - u)^{7}$.
$t^{7}-7t^{6}u + 21t^{5}u^{2}-35t^{4}u^{3}+35t^{3}u^{4}+\square+7tu^{6}-u^{7}$

Explanation:

Step1: Recall binomial expansion formula

The binomial expansion of \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\). For \((t-u)^7=(t+(-u))^7\), the general term is \(T_{r + 1}=\binom{7}{r}t^{7 - r}(-u)^{r}\).

Step2: Find the missing term

The powers of \(t\) and \(u\) follow the pattern: for the first term \(t^7\) (\(r = 0\)), then \(t^6u\) (\(r=1\)), \(t^5u^2\) (\(r = 2\)), \(t^4u^3\) (\(r=3\)), \(t^3u^4\) (\(r = 4\)), the next term should have \(r = 5\).
Using the formula \(T_{6}=\binom{7}{5}t^{7-5}(-u)^{5}\).
Since \(\binom{7}{5}=\binom{7}{2}=\frac{7!}{2!(7 - 2)!}=\frac{7\times6}{2\times 1}=21\), and \((-u)^{5}=-u^{5}\), \(t^{7 - 5}=t^{2}\).
So \(T_{6}=21t^{2}(-u)^{5}=- 21t^{2}u^{5}\)

Answer:

\(-21t^{2}u^{5}\)