To expand the expression [tex]\((2x^3 + 3y^2)^7\)[/tex] using the binomial theorem, we need Pascal's triangle with 8 rows. This is because, for an expansion involving the power [tex]\(n\)[/tex], Pascal's triangle requires [tex]\(n + 1\)[/tex] rows. In this case, [tex]\(n = 7\)[/tex], so we need 7 + 1 = 8 rows.
Here is Pascal's triangle with 8 rows:
[tex]\[
\begin{array}{ccccccccccccccccccccccccccccccccccccccc}
& & & & & & & & 1 & & & & & & & & \\
& & & & & & & 1 & & 1 & & & & & & \\
& & & & & & 1 & & 2 & & 1 & & & & & \\
& & & & & 1 & & 3 & & 3 & & 1 & & & & & \\
& & & & 1 & & 4 & & 6 & & 4 & & 1 & & & & \\
& & & 1 & & 5 & & 10 & & 10 & & 5 & & 1 & & \\
& & 1 & & 6 & & 15 & & 20 & & 15 & & 6 & & 1 & \\
& 1 & & 7 & & 21 & & 35 & & 35 & & 21 & & 7 & & 1 \\
\end{array}
\][/tex]
Describing the rows:
- Row 0: [tex]\([1]\)[/tex]
- Row 1: [tex]\([1, 1]\)[/tex]
- Row 2: [tex]\([1, 2, 1]\)[/tex]
- Row 3: [tex]\([1, 3, 3, 1]\)[/tex]
- Row 4: [tex]\([1, 4, 6, 4, 1]\)[/tex]
- Row 5: [tex]\([1, 5, 10, 10, 5, 1]\)[/tex]
- Row 6: [tex]\([1, 6, 15, 20, 15, 6, 1]\)[/tex]
- Row 7: [tex]\([1, 7, 21, 35, 35, 21, 7, 1]\)[/tex]
These coefficients will be used to expand the given expression [tex]\((2x^3 + 3y^2)^7\)[/tex] according to the binomial theorem formula:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
In this case, [tex]\(a = 2x^3\)[/tex] and [tex]\(b = 3y^2\)[/tex].
Thus, each term in the expansion can be determined using the coefficients from Pascal's triangle, and the general term will be:
[tex]\[
\text{term}_k = \binom{7}{k} (2x^3)^{7-k} (3y^2)^k
\][/tex]
Where [tex]\(k\)[/tex] ranges from 0 to 7.
