Find The 6th Term In $(a+b)^9$ Expansion

Hey guys! Ever wondered how to crack those binomial expansion problems? Today, we're diving deep into finding the value of the 6th term in the expansion of $(a+b)^9$. This is a classic math problem that pops up a lot, so mastering it will seriously boost your skills. We'll break down the binomial theorem and walk you through each step, making sure you’ve got this down pat. Get ready to flex those math muscles!

Understanding the Binomial Theorem

The binomial theorem is your best friend when dealing with expressions like $(a+b)^n$. It gives us a formula to expand such expressions without actually multiplying everything out. The general formula looks like this:

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

Here, $\binom{n}{k}$ is the binomial coefficient, often read as "n choose k", and it's calculated as $\frac{n!}{k!(n-k)!}$. The 'n' is the power to which we're raising our binomial, and 'k' is the index of the term we're interested in, starting from 0. So, the first term corresponds to k=0, the second to k=1, and so on. This means the $(k+1)^{th}$ term in the expansion is given by $\binom{n}{k} a^{n-k} b^k$. This formula is super powerful because it lets us pinpoint any term in the expansion directly. When we’re asked for the 6th term, remember that our 'k' value will be 5, because we start counting from k=0. So, the 6th term is actually the term where k=5. Understanding this index shift is crucial for getting the right answer, so always double-check if you're looking for the $m^{th}$ term, you'll use $k = m-1$. This theorem is fundamental in algebra and has applications in probability, calculus, and beyond. It’s a foundational concept that opens doors to more complex mathematical ideas.

Calculating the 6th Term

Alright, let's get down to business and calculate the value of the 6th term for our specific problem, which is $(a+b)^9$. Here, our 'n' is 9. Since we want the 6th term, we know that $k = 6 - 1 = 5$. Now, we plug these values into the general term formula: $\binom{n}{k} a^{n-k} b^k$.

Substituting $n=9$ and $k=5$, we get:

$$\binom{9}{5} a^{9-5} b^5$$

This simplifies to:

$$\binom{9}{5} a^4 b^5$$

Now, the critical part is calculating the binomial coefficient $\binom{9}{5}$. Remember the formula: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

So, for $\binom{9}{5}$, we have:

$$\binom{9}{5} = \frac{9!}{5!(9-5)!} = \frac{9!}{5!4!} = \frac{9 \times 8 \times 7 \times 6 \times 5!}{5! \times (4 \times 3 \times 2 \times 1)}$$

We can cancel out the $5!$ from the numerator and denominator:

$$\binom{9}{5} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}$$

Let's simplify this calculation. The denominator is $4 \times 3 \times 2 \times 1 = 24$. The numerator is $9 \times 8 \times 7 \times 6$. We can do some cancellations to make it easier:

• $8 / (4 \times 2) = 1$
• $9 / 3 = 3$
• $6$ remains.

So, we have $3 \times 1 \times 7 \times 6 = 126$.

Therefore, the binomial coefficient $\binom{9}{5}$ is 126.

Putting it all together, the 6th term of the expansion is $126 a^4 b^5$. This involves careful calculation of factorials and coefficients, but breaking it down step-by-step makes it manageable. Remember, practice is key to becoming comfortable with these calculations, and understanding the underlying formula helps immensely.

Analyzing the Options

We've worked through the calculation and found that the value of the 6th term in the expansion of $(a+b)^9$ is $126 a^4 b^5$. Now, let's look at the options provided to see which one matches our result.

A. $84 a^3 b^6$ B. $84 a^6 b^3$ C. $126 a^4 b^5$ D. $126 a^5 b^4$

Comparing our calculated term, $126 a^4 b^5$, with the options, we can see that option C is an exact match.

Let's quickly check why the other options are incorrect. Options A and B have a coefficient of 84, which we didn't get from our binomial coefficient calculation $\binom{9}{5}$. This coefficient is determined by 'n' and 'k', so if it's different, the term is wrong. Options A and B also have different powers for 'a' and 'b'. Remember, the powers must add up to 'n' (which is 9 in this case), and follow the pattern $a^{n-k}b^k$. For option A, $3+6=9$, but the coefficient is wrong. For option B, $6+3=9$, but the coefficient is wrong.

Option D has the correct coefficient, 126, but the powers are $a^5 b^4$. This means that $n-k=5$ and $k=4$. If $k=4$, then $n-k = 9-4=5$, which gives $a^5b^4$. However, we were looking for the 6th term, which corresponds to $k=5$, not $k=4$. So, option D represents the 5th term ($k=4$) of the expansion, not the 6th term. This highlights the importance of correctly identifying the value of 'k' based on the term number. Our calculation of $126 a^4 b^5$ for the 6th term (where $k=5$) is definitely correct. It's easy to get mixed up with the powers or the term number, so always double-check your work against the binomial theorem's structure.

Why This Matters

Mastering problems like finding the value of the 6th term in an expansion isn't just about acing a test; it's about building a solid foundation in algebra. The binomial theorem is a fundamental concept used in various fields, from statistics and probability to advanced calculus and computer science. For instance, in probability, it's used to calculate probabilities in scenarios involving a fixed number of independent trials, like coin flips or success/failure events. In statistics, it helps in understanding distributions like the binomial distribution.

Knowing how to efficiently expand binomials and identify specific terms saves a ton of time compared to brute-force multiplication, especially when 'n' is large. This skill is also crucial for understanding more complex mathematical concepts like Taylor series expansions, which are used to approximate functions. So, when you're working through these problems, remember you're not just solving for 'x' or finding a specific term; you're sharpening tools that are essential for higher-level mathematics and scientific applications. Keep practicing, guys, and you'll find these concepts becoming second nature!