Solving For N In A Binomial Summation 3(n C 1) + 3^2(n C 2) + 3^n(n C N) = 4095

Hey guys! Ever stumbled upon a math problem that looks like it's written in another language? Well, let's dive into one together and break it down. We're going to tackle this equation:

3 * (n C 1) + 3^2 * (n C 2) + ... + 3^n * (n C n) = 4095

And our mission, should we choose to accept it, is to find the positive integer value of 'n' that makes this equation true. Buckle up, because we're about to embark on a mathematical adventure!

Decoding the Equation: A Journey into Binomial Coefficients

Before we jump into solving, let's decode what this equation is actually saying. The heart of the equation lies in the binomial coefficients, represented as (n C k). If you're scratching your head, don't worry! Let's break it down.

Binomial coefficients, often read as "n choose k", tell us how many ways we can select 'k' items from a set of 'n' items, without considering the order. Think of it like this: if you have 'n' friends and you want to form a team of 'k' people, (n C k) tells you how many different teams you can make.

The formula for calculating binomial coefficients is:

(n C k) = n! / (k! * (n-k)!)

Where "!" denotes the factorial, meaning the product of all positive integers up to that number. For example, 5! = 5 * 4 * 3 * 2 * 1 = 120.

So, in our equation, each term involves a binomial coefficient multiplied by a power of 3. Let's look at the first few terms to get a better feel:

• 3 * (n C 1): This is 3 times the number of ways to choose 1 item from 'n' items.
• 3^2 * (n C 2): This is 9 times the number of ways to choose 2 items from 'n' items.
• And so on, until we reach 3^n * (n C n), which is 3^n times the number of ways to choose 'n' items from 'n' items (which is always 1).

Now that we understand the building blocks, we can see that the equation is a sum of these terms, and we need to find the 'n' that makes this sum equal to 4095.

The Binomial Theorem: Our Secret Weapon

Okay, guys, here's where things get really interesting. We're going to pull out a mathematical tool called the Binomial Theorem. This theorem provides a powerful shortcut for expanding expressions of the form (x + y)^n. It states:

(x + y)^n = (n C 0) * x^n * y^0 + (n C 1) * x^(n-1) * y^1 + (n C 2) * x^(n-2) * y^2 + ... + (n C n) * x^0 * y^n

Notice anything familiar? The binomial coefficients are front and center! This theorem looks complicated, but it's a perfect match for our equation. Let's see why.

If we set x = 1 and y = 3 in the Binomial Theorem, we get:

(1 + 3)^n = (n C 0) * 1^n * 3^0 + (n C 1) * 1^(n-1) * 3^1 + (n C 2) * 1^(n-2) * 3^2 + ... + (n C n) * 1^0 * 3^n

Simplifying this, we have:

4^n = (n C 0) + 3 * (n C 1) + 3^2 * (n C 2) + ... + 3^n * (n C n)

This looks incredibly similar to our original equation! The only difference is the (n C 0) term on the right side. Remember that (n C 0) is the number of ways to choose 0 items from 'n' items, which is always 1. So, we can rewrite the equation as:

4^n = 1 + 3 * (n C 1) + 3^2 * (n C 2) + ... + 3^n * (n C n)

Now, let's rearrange our original equation:

3 * (n C 1) + 3^2 * (n C 2) + ... + 3^n * (n C n) = 4095

Adding 1 to both sides, we get:

1 + 3 * (n C 1) + 3^2 * (n C 2) + ... + 3^n * (n C n) = 4096

Boom! We've found the connection. The left side of this equation is exactly the same as the right side of our Binomial Theorem expansion. This means:

4^n = 4096

Cracking the Code: Finding the Value of n

We've transformed a seemingly complex equation into a much simpler one: 4^n = 4096. Now, the challenge is to find the integer 'n' that satisfies this equation.

To do this, we need to express 4096 as a power of 4. Let's try some powers of 4:

• 4^1 = 4
• 4^2 = 16
• 4^3 = 64
• 4^4 = 256
• 4^5 = 1024
• 4^6 = 4096

Eureka! We found it. 4096 is equal to 4^6. Therefore, our equation becomes:

4^n = 4^6

For this equation to hold true, the exponents must be equal. So, we have:

n = 6

And that's it! We've successfully found the positive integer value of 'n' that satisfies the original equation. The answer is n = 6.

Wrapping Up: A Mathematical Triumph

Guys, we did it! We took on a challenging problem involving binomial coefficients and summations, and we conquered it using the Binomial Theorem. This problem highlights the power of mathematical tools and how they can help us simplify complex expressions.

Remember, the key to solving these kinds of problems is to break them down into smaller, manageable parts. Understand the definitions, look for patterns, and don't be afraid to use the tools in your mathematical arsenal. And most importantly, have fun with it!

So, the next time you encounter a seemingly daunting equation, remember our journey today. You might just surprise yourself with what you can achieve. Keep exploring, keep learning, and keep those mathematical gears turning!

Choosing the Correct Answer and Final Thoughts

Going back to the original question, we were asked to choose the correct value of 'n' from the given options. We found that n = 6, which corresponds to option C.

Therefore, the correct answer is:

• C) 6

In conclusion, this problem beautifully illustrates how the Binomial Theorem can be applied to solve seemingly complex summation problems. By recognizing the pattern and applying the theorem, we were able to simplify the equation and efficiently find the solution. Keep practicing and exploring these concepts, guys, and you'll become mathematical masters in no time!