Binomial Expansion: Choosing 'k' Values Explained

Hey guys! Let's dive into a binomial expansion problem that Kerys is tackling. She's trying to expand the binomial expression $(2x + y)^4$ using the binomial theorem. The binomial theorem expression she's using is $\sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$. The big question is: what values should she substitute for k, and, just as importantly, why?

Understanding the Binomial Theorem

Before we jump into the specifics, let's break down the binomial theorem a bit. This theorem is a powerful tool for expanding expressions of the form $(a + b)^n$, where n is a non-negative integer. It tells us exactly how to expand this expression into a sum of terms, each involving powers of a and b. The coefficients of these terms are binomial coefficients, which you might recognize as the numbers in Pascal's Triangle.

The Formula:

The general formula for the binomial theorem is:

$(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$

Where:

• n is the power to which the binomial is raised.
• k is an index that ranges from 0 to n. It determines the power of b in each term and helps calculate the binomial coefficient.
• $\binom{n}{k}$ is the binomial coefficient, also written as "n choose k", and it's calculated as $\frac{n!}{k!(n-k)!}$, where "!" denotes the factorial.
• a and b are the terms within the binomial.

Why It Works:

The binomial theorem works because it systematically accounts for every possible combination of choosing a or b from each of the n factors in the expansion of $(a + b)^n$. Each term in the expansion corresponds to a specific combination, and the binomial coefficient tells us how many times that combination occurs.

Breaking Down Kerys's Problem

In Kerys's case, she's dealing with $(2x + y)^4$. So, let's identify the components:

• a = 2x
• b = y
• n = 4

Now, the question is what values to use for k. The summation notation $\sum_{k=0}^n$ tells us that k starts at 0 and goes up to n. Since n is 4 in this problem, k will take on the values 0, 1, 2, 3, and 4. Let's see why these values are crucial.

Why These Values of k?

k is the engine that drives the binomial theorem, dictating the specific terms that emerge in the expansion. By incrementing k from 0 to n, we ensure that every possible combination of a and b is accounted for, giving us the complete expansion.

Let's walk through each value of k to see its effect:

• k = 0:

  • Term: $\binom{4}{0} (2x)^{4-0} y^0 = \binom{4}{0} (2x)^4 y^0 = 1 \cdot 16x^4

  This term gives us the first term in the expansion. Notice how it includes (2x) raised to the highest power (4) and y raised to the power of 0 (which is just 1). This is where we start, guys! It sets the stage for the rest of the expansion.

• k = 1:

  • Term: $\binom{4}{1} (2x)^{4-1} y^1 = \binom{4}{1} (2x)^3 y^1 = 4

  Here, the power of (2x) decreases by one, and the power of y increases by one. The binomial coefficient $\binom{4}{1}$ tells us how many ways we can choose one 'y' from the four factors of (2x + y). Keep an eye on those coefficients; they're super important.

• k = 2:

  • Term: $\binom{4}{2} (2x)^{4-2} y^2 = \binom{4}{2} (2x)^2 y^2 = 6

  Now, we have (2x) squared and y squared. $\binom{4}{2}$ tells us the number of ways to pick two 'y's from the four factors. The pattern is becoming clearer, right?

• k = 3:

  • Term: $\binom{4}{3} (2x)^{4-3} y^3 = \binom{4}{3} (2x)^1 y^3 = 4

  The power of (2x) is now 1, and the power of y is 3. $\binom{4}{3}$ gives the number of ways to choose three 'y's. See how the symmetry is starting to show?

• k = 4:

  • Term: $\binom{4}{4} (2x)^{4-4} y^4 = \binom{4}{4} (2x)^0 y^4 = 1

  Finally, (2x) is raised to the power of 0 (which is 1), and y is raised to the power of 4. This is the last term in the expansion.

By plugging in each value of k from 0 to 4, we systematically generate all the terms in the expansion of $(2x + y)^4$. Each k value corresponds to a unique term with a specific combination of powers of (2x) and y, and the binomial coefficient tells us how many times each combination appears.

Putting It All Together: The Expanded Form

Now that we've explored each value of k, let's assemble the complete expansion:

$(2x + y)^4 = \binom{4}{0} (2x)^4 y^0 + \binom{4}{1} (2x)^3 y^1 + \binom{4}{2} (2x)^2 y^2 + \binom{4}{3} (2x)^1 y^3 + \binom{4}{4} (2x)^0 y^4$

Calculate the binomial coefficients and simplify:

$(2x + y)^4 = 1(16x^4)(1) + 4(8x^3)(y) + 6(4x^2)(y^2) + 4(2x)(y^3) + 1(1)(y^4)$

$(2x + y)^4 = 16x^4 + 32x^3y + 24x^2y^2 + 8xy^3 + y^4$

So, the complete expansion of $(2x + y)^4$ is $16x^4 + 32x^3y + 24x^2y^2 + 8xy^3 + y^4$.

Kerys's Takeaway

Kerys should substitute the values 0, 1, 2, 3, and 4 for k in the binomial theorem expression. This is because these values systematically cover all possible combinations of the terms (2x) and y in the expansion of $(2x + y)^4$. Each value of k generates a unique term in the expansion, ensuring that no term is missed. By understanding this, Kerys (and you!) can confidently tackle any binomial expansion problem!

In summary, k is not just some arbitrary index; it’s the key to unlocking the structure of the binomial expansion. By stepping k through each integer value from 0 to n, we systematically generate and account for every possible term in the expansion, with the binomial coefficients providing the correct numerical weight to each. So, next time you see a binomial expansion, remember the power of k! It's like the secret ingredient in a recipe for polynomial success! You got this!