To expand [tex]\((x^{2L} + 1)^4\)[/tex] using Pascal's Triangle, follow these steps:
### Step 1: Coefficients from Pascal's Triangle
First, recognize that the coefficients for the expansion come from the 4th row of Pascal's Triangle. The coefficients for [tex]\((a + b)^4\)[/tex] are [tex]\(1, 4, 6, 4, 1\)[/tex].
### Step 2: Apply the Binomial Theorem
The Binomial Theorem states:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
In our case, [tex]\(a = x^{2L}\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(n = 4\)[/tex]. So now we will plug these into the binomial expansion formula.
### Step 3: Expand Each Term
Using the coefficients from Pascal's Triangle and the binomial theorem, we get:
[tex]\[
(x^{2L} + 1)^4 = \sum_{k=0}^{4} \binom{4}{k} (x^{2L})^{4-k} (1)^k
\][/tex]
### Step 4: Calculate Each Term
Now we can break this down into individual terms:
1. For [tex]\(k = 0\)[/tex]:
[tex]\[ \binom{4}{0} (x^{2L})^{4-0} (1)^0 = 1 \cdot x^{8L} \cdot 1 = x^{8L} \][/tex]
2. For [tex]\(k = 1\)[/tex]:
[tex]\[ \binom{4}{1} (x^{2L})^{4-1} (1)^1 = 4 \cdot x^{6L} \cdot 1 = 4x^{6L} \][/tex]
3. For [tex]\(k = 2\)[/tex]:
[tex]\[ \binom{4}{2} (x^{2L})^{4-2} (1)^2 = 6 \cdot x^{4L} \cdot 1 = 6x^{4L} \][/tex]
4. For [tex]\(k = 3\)[/tex]:
[tex]\[ \binom{4}{3} (x^{2L})^{4-3} (1)^3 = 4 \cdot x^{2L} \cdot 1 = 4x^{2L} \][/tex]
5. For [tex]\(k = 4\)[/tex]:
[tex]\[ \binom{4}{4} (x^{2L})^{4-4} (1)^4 = 1 \cdot x^{0} \cdot 1 = 1 \][/tex]
### Step 5: Combine All Terms
Now, sum all the terms together:
[tex]\[
(x^{2L} + 1)^4 = x^{8L} + 4x^{6L} + 6x^{4L} + 4x^{2L} + 1
\][/tex]
### Final Answer
The expanded form of [tex]\((x^{2L} + 1)^4\)[/tex] is:
[tex]\[
x^{8L} + 4x^{6L} + 6x^{4L} + 4x^{2L} + 1
\][/tex]
