Question

1. use pascals triangle to expand the binomial $(x - 1)^6$.

$x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1$
$x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x - 1$
$x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1$
$x^6 + 6x^5 + 15x^4 + 20x^3 - 15x^2 - 6x - 1$

Explanation:

Step1: Get Pascal's Triangle row for $n=6$

The 6th row (starting from row 0) of Pascal's Triangle is $1, 6, 15, 20, 15, 6, 1$.

Step2: Apply binomial expansion formula

For $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k}a^{n-k}b^k$, here $a=x$, $b=-1$, $n=6$.

$$\begin{align*} (x-1)^6 &= \binom{6}{0}x^6(-1)^0 + \binom{6}{1}x^5(-1)^1 + \binom{6}{2}x^4(-1)^2 + \binom{6}{3}x^3(-1)^3 + \binom{6}{4}x^2(-1)^4 + \binom{6}{5}x^1(-1)^5 + \binom{6}{6}x^0(-1)^6 \end{align*}$$

Step3: Calculate each term

• Term1: $1 \cdot x^6 \cdot 1 = x^6$
• Term2: $6 \cdot x^5 \cdot (-1) = -6x^5$
• Term3: $15 \cdot x^4 \cdot 1 = 15x^4$
• Term4: $20 \cdot x^3 \cdot (-1) = -20x^3$
• Term5: $15 \cdot x^2 \cdot 1 = 15x^2$
• Term6: $6 \cdot x \cdot (-1) = -6x$
• Term7: $1 \cdot 1 \cdot 1 = 1$

Step4: Combine all terms

Add the calculated terms together.

Answer:

$x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1$ (the third option)