Question

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 $1s^3 + 3s^2t^1 + 3st^2 + 1t^3$

Explanation:

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

Pascal's Triangle for the 3rd power (\(n = 3\)) has coefficients \(1, 3, 3, 1\). The binomial expansion of \((a + b)^n\) is \(\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(\binom{n}{k}\) are the Pascal's Triangle coefficients. For \((s + t)^3\), \(a = s\), \(b = t\), \(n = 3\).

Step2: Apply the binomial expansion formula

Using the coefficients \(1, 3, 3, 1\), the expansion is:

$$
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$$

Looking at the options, option e is \(1s^{3}+3s^{2}t^{1}+3st^{2}+1t^{3}\) (which is equivalent to the expansion we derived, with the coefficients and exponents matching).

Answer:

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