Question

unit 5 test day 1
name:

1. circle the best answer:

use pascals triangle to write the binomial expansion of:
$(s + t)^3$

a $3s + 3t$
b $1s^3t + 3s^2t^1 + 3st^2 + 1st^3$
c $4s^3 + 6s^2t^1 + 6st^2 + 1t^3$
d $1s^3t^3 + 3s^2t^2 + 3st + 1$
e $1s^3 + 3s^2t^1 + 3st^2 + 1t^3$

1. circle the best answer:

use pascals triangle to write the binomial expansion of:
$(c + d)^4$

a $1c^4 + 3c^3d^1 + 9c^2d^2 + 3c^1d^3 + 1d^4$
b $1c^4 + 4c^3d^1 + 6c^2d^2 + 4c^1d^3 + 1d^4$
c $1c^4d + 4c^3d^1 + 6c^2d^2 + 4c^1d^3 + 1cd^4$
d $4c + 4d$
e $c^4 + c^3d^1 + c^2d^2 + c^1d^3 + d^4$

Explanation:

Question 1

Step1: Recall Pascal's Triangle for \(n = 3\)

Pascal's Triangle for \(n = 3\) (the exponent in \((s + t)^3\)) has coefficients \(1, 3, 3, 1\).

Step2: Apply Binomial Expansion Formula

The binomial expansion of \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\). For \((s + t)^3\), \(a=s\), \(b = t\), \(n = 3\). Using the coefficients from Pascal's Triangle (\(1,3,3,1\)):

• When \(k = 0\): \(1\times s^{3}t^{0}=s^{3}\)
• When \(k = 1\): \(3\times s^{2}t^{1}=3s^{2}t\)
• When \(k = 2\): \(3\times s^{1}t^{2}=3st^{2}\)
• When \(k = 3\): \(1\times s^{0}t^{3}=t^{3}\)

So the expansion is \(1s^{3}+3s^{2}t + 3st^{2}+1t^{3}\) (option e).

Step1: Recall Pascal's Triangle for \(n = 4\)

Pascal's Triangle for \(n = 4\) has coefficients \(1,4,6,4,1\).

Step2: Apply Binomial Expansion Formula

For \((c + d)^4\), \(a = c\), \(b=d\), \(n = 4\). Using the coefficients from Pascal's Triangle (\(1,4,6,4,1\)):

• When \(k = 0\): \(1\times c^{4}d^{0}=c^{4}\)
• When \(k = 1\): \(4\times c^{3}d^{1}=4c^{3}d\)
• When \(k = 2\): \(6\times c^{2}d^{2}=6c^{2}d^{2}\)
• When \(k = 3\): \(4\times c^{1}d^{3}=4cd^{3}\)
• When \(k = 4\): \(1\times c^{0}d^{4}=d^{4}\)

So the expansion is \(1c^{4}+4c^{3}d+6c^{2}d^{2}+4cd^{3}+1d^{4}\) (option b).

Answer:

e. \(1s^{3}+3s^{2}t^{1}+3st^{2}+1t^{3}\)

Question 2