Question

Expand \((1+x)^{8}\) using the Binomial Theorem.

Short answer

The expansion is \((1 + x)^8 = 1 + 8x + 28x^2 + 56x^3 + 70x^4 + 56x^5 + 28x^6 + 8x^7 + x^8\).

Step-by-step solution

Understand the Binomial Theorem

The Binomial Theorem states that \[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]where \(\binom{n}{k}\) is the binomial coefficient and represents the number of ways to choose \(k\) elements from \(n\) elements, given by \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\). Here, \(a = 1\), \(b = x\), and \(n = 8\).

Write the Expansion Formula

Plug in the values \(a = 1\), \(b = x\), and \(n = 8\) into the Binomial Theorem to get:\[(1 + x)^8 = \sum_{k=0}^{8} \binom{8}{k} (1)^{8-k} x^k\]Since \((1)^{8-k} = 1\) for all \(k\), the expression simplifies to:\[(1 + x)^8 = \sum_{k=0}^{8} \binom{8}{k} x^k\]

Calculate Binomial Coefficients

Calculate the binomial coefficients \(\binom{8}{k}\) for \(k = 0, 1, 2, \ldots, 8\).These are:- \(\binom{8}{0} = 1\)- \(\binom{8}{1} = 8\)- \(\binom{8}{2} = 28\)- \(\binom{8}{3} = 56\)- \(\binom{8}{4} = 70\)- \(\binom{8}{5} = 56\)- \(\binom{8}{6} = 28\)- \(\binom{8}{7} = 8\)- \(\binom{8}{8} = 1\)

Write the Expanded Form

Substitute the coefficients from Step 3 into the expansion:\[(1 + x)^8 = \binom{8}{0} x^0 + \binom{8}{1} x^1 + \binom{8}{2} x^2 + \binom{8}{3} x^3 + \binom{8}{4} x^4 + \binom{8}{5} x^5 + \binom{8}{6} x^6 + \binom{8}{7} x^7 + \binom{8}{8} x^8\]This equals:\[(1 + x)^8 = 1 + 8x + 28x^2 + 56x^3 + 70x^4 + 56x^5 + 28x^6 + 8x^7 + x^8\]

Key concepts

Binomial Coefficients

To understand the process of expanding binomials, it's important to grasp the concept of binomial coefficients. Binomial coefficients are a key part of the Binomial Theorem. They indicate the number of ways to choose a subset of elements from a larger set. If we have a set of size \(n\) and want to pick \(k\) items from it, the binomial coefficient \(\binom{n}{k}\) gives us that count.

These coefficients can be calculated using the formula:

• \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

where \(!\) denotes factorial, meaning the product of all positive integers up to that number.

For instance, calculating \(\binom{8}{3}\) involves:

• \(\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\)

This number tells us there are 56 ways to pick 3 objects from a total of 8. Alongside the values you saw in the solution, these provide the weights for each term in the polynomial expansion.

Polynomial Expansion

Polynomial expansion using the Binomial Theorem allows us to express a power of a binomial like \((a + b)^n\) as a sum of terms. Each term corresponds to a different power of \(b\) multiplied by a binomial coefficient.

The general form is:

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

In this formula, each term in the sum features different powers of \(a\) and \(b\), gradually increasing \(b\) from \(b^0\) to \(b^n\) while reducing \(a\) by the same steps.

In the exercise you practiced with, we used \(a = 1\) and \(b = x\) for simplicity. Because anything raised to the power of another is 1, each term simply became \(\binom{8}{k}x^k\). This means it's just about calculating different powers of \(x\) and assigning them their respective binomial coefficients.

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arranging, and grouping items. Many tools in combinatorics, like the binomial coefficient, help solve problems involving possible selections or arrangements.

The exercise illustrates combinatorics by showing how to calculate the different possible combinations of items (or terms in the expansion). The reason \(\binom{n}{k}\) works in the Binomial Theorem is due to its combinatorial nature: it considers all different ways to select \(k\) items from a total of \(n\).

If thinking of the problem as picking apples and oranges helps, expand the ideas:

• Consider \((1 + x)^8\). It's as if we are considering all distributions of 8 identical "items" (like numbered slots) between two categories (one's slots and \(x\)'s slots).
• Each \(\binom{n}{k}\) is like asking "How many ways can I distribute these?"

This tangible perspective makes the idea of binomial coefficients incredibly powerful and versatile for solving just about any problem involving selection or arrangement.