Question

Write the requested term of each binomial expansion, and simplify. Ninth term of \(\left(x^{2}+1\right)^{15}\)

Short answer

The ninth term of \((x^2 + 1)^{15}\) is \(6435x^{14}\).

Step-by-step solution

Understand the Binomial Theorem

The Binomial Theorem states that the expansion of \(a + b)^n\) can be expressed as \(\sum_{k=0}^{n}{\binom{n}{k}a^{n-k}b^{k}}\). Each term in the expansion can be calculated using the combination function: the k-th term is given by \(\binom{n}{k}a^{n-k}b^{k}\).

Identify Parameters for the Specific Term

We are looking for the ninth term of \(\left(x^{2}+1\right)^{15}\). This means that \((n = 15)\) and \((a = x^2)\), \((b = 1)\), and we want \((k = 9 - 1 = 8)\), since the first term corresponds to \((k = 0)\).

Calculate the Binomial Coefficient

Using the combination formula, we calculate \(\binom{15}{8} = \frac{15!}{8!(15 - 8)!}\). Simplify this to get the coefficient.

Write Out the Term

Now that we have the coefficient, we write out the term: \(\binom{15}{8}(x^2)^{15 - 8}(1)^8\).

Simplify the Term

Simplify the term to find the actual ninth term of the expansion. Since \(1^8 = 1\), it doesn't change the value, so the term simplifies to \(\binom{15}{8}x^{2*(15 - 8)}\) which equals \(\binom{15}{8}x^{14}\).

Calculate the Numerical Value

Calculate \(\binom{15}{8}\) which equals \(6435\). Therefore, the complete ninth term is \(6435x^{14}\).

Key concepts

Binomial Coefficient

The binomial coefficient is a fundamental component when dealing with the Binomial Theorem. It's represented as \(\binom{n}{k}\), where \(n\) is the power to which the binomial is raised, and \(k\) is the specific term we are interested in. It essentially determines how many ways \(k\) items can be chosen from a set of \(n\) items without considering the order of selection.

In the context of the ninth term of the binomial expansion of \(\left(x^{2}+1\right)^{15}\), the binomial coefficient is responsible for giving us the multiplier of that term. For instance, in this specific exercise, we'd be looking at \(\binom{15}{8}\). This acts as a magnifying glass, highlighting not just the ninth term but the importance of each term in the expansion. As we journey through our binomial expansion, we need to remember that this magical number isn't just a random occurrence – it's the result of a well-defined mathematical concept that has applications far beyond just this theorem – from statistics to combinatorics!

Combination Formula

When you hear about the combination formula, envision a puzzle where you're tasked to find the total number of unique groupings possible. In mathematical terms, this formula is used to calculate the binomial coefficient and is expressed as \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), where \(n!\) denotes factorial of \(n\), nothing but a shorthand for multiplying a series of descending natural numbers.

If we delve into our textbook problem, applying the combination formula to determine the ninth term of \(\left(x^{2}+1\right)^{15}\), we use \(n = 15\) and \(k = 8\), resulting in a binomial coefficient of \(\binom{15}{8}\). The combination formula is the unsung hero here – it meticulously assembles the puzzle, showing us exactly how many ways we can select 8 items out of 15, providing the cornerstone for our entire binomial expansion process.

Binomial Expansion Term Simplification

Once we have grasped the binomial coefficient and the combination formula, we can navigate to the final stage: binomial expansion term simplification. It’s like reducing a complex recipe to its essential ingredients. Here, the Binomial Theorem offers the blueprint for constructing each term of our expansion. But it’s simplification that clears away the clutter to reveal the term's simplest form.

This is particularly evident in our ninth term for the binomial \(\left(x^{2}+1\right)^{15}\). After determining the binomial coefficient \(6435\), we’re left with \(6435(x^2)^{15-8}\). A glance reveals redundant units like \(1^8\), which simplifies to \(1\), essentially not impacting the term at all. Thus, simplification hones in on the purest expression of our target term, \(6435x^{14}\), slicing through mathematical complexity with a satisfying snap.