Cracking the Code: How the Binomial Theorem Solves a Futuristic Farming Mystery

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Ever looked at a complex system and wondered how to make sense of it all? Let’s jump into a not-so-distant future. Imagine a bustling metropolis with skyscrapers that double as high-tech vertical farms. A team of brilliant botanists is working on a new super-herb, and its yield stability is governed by a surprisingly familiar mathematical model: $(x – 3y)^7$.

But there’s a problem. The head botanist has found two strange anomalies. First, a ‘stable yield’ configuration that seems completely unaffected by light intensity (‘x’). What is this constant value? Second, a ‘peak performance’ mode is directly tied to the $x^3$ factor in their system. What’s the precise impact, or coefficient, of this setup? To crack this code, we don’t need a futuristic computer; we just need one of the most powerful tools in our IB Math toolkit: the Binomial Theorem. Let’s get to it!

Table of Contents

What is the Binomial Theorem?

Before we can be heroes in the vertical farm, we need to understand our primary tool. The Binomial Theorem is essentially a recipe for expanding expressions of the form $(a+b)^n$ without having to do the tedious, repetitive multiplication. It’s a lifesaver, especially when ‘n’ gets large.

The General Formula

Rule: The Binomial Theorem

For any binomial $(a+b)$ raised to a non-negative integer power $n$, the expansion is given by:

$(a+b)^n = \sum\limits_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$

Which expands to:

$(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1}b^1 + \dots + \binom{n}{n}a^0 b^n$

The term $\binom{n}{r}$ is the binomial coefficient, read as ‘n choose r’, and it’s calculated as $\frac{n!}{r!(n-r)!}$. The good news? This formula is right there in your IB Math Formula Booklet!

The ‘General Term’ – Your Secret Weapon

Let’s be real, you’ll rarely be asked to write out the entire expansion of something like $(x – 3y)^7$. That would take forever! The real power for IB exams comes from isolating a single piece of that expansion. This is where the ‘General Term’ formula comes in.

Rule: The General Term

The $(r+1)^{th}$ term of the expansion of $(a+b)^n$ is given by:

$T_{r+1} = \binom{n}{r} a^{n-r} b^r$

Watch Out for ‘r’!

This is the #1 spot for silly mistakes. Notice the term is $T_{r+1}$. The ‘r’ value in the formula always starts at 0. So:

  • For the 1st term, you use $r=0$.
  • For the 5th term, you use $r=4$.
  • For the 10th term, you use $r=9$.

Always remember that the term number is one greater than the value of ‘r’ you plug into the formula!

Applying the Binomial Theorem: Core Skills

Alright, let’s get our hands dirty. There are three main types of questions you’ll see involving the general term. Master these, and you’re well on your way.

Finding a Specific Term

This is the most straightforward application. You’re asked for a specific term in the sequence, like the 3rd or 5th term.

Example: Finding the 4th term

Problem: Find the 4th term in the expansion of $(2p + q)^8$.

Step 1: Identify your components.

We’re using the formula $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.

From $(2p + q)^8$, we have: $a = 2p$, $b = q$, and $n=8$.

Step 2: Find the value of ‘r’.

We want the 4th term, so $r+1=4$. This means we need to use $r=3$.

Step 3: Substitute and solve.

$T_{3+1} = T_4 = \binom{8}{3} (2p)^{8-3} (q)^3$

$T_4 = \binom{8}{3} (2p)^5 (q)^3$

First, calculate the binomial coefficient: $\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = 56$.

Next, handle the powers: $(2p)^5 = 2^5 p^5 = 32p^5$.

Now combine everything: $T_4 = 56 \cdot (32p^5) \cdot (q)^3 = 1792p^5q^3$.

Answer: The 4th term is $1792p^5q^3$.

Finding the Coefficient of a Specific Power

This is a subtle variation. Here, you’re not asked for the whole term, just the number in front of a specific variable part, like the coefficient of $y^4$.

Example: Finding a coefficient

Problem: Find the coefficient of the $y^4$ term in the expansion of $(y – 5)^7$.

Step 1: Identify components.

$a = y$, $b = -5$ (don’t forget the negative sign!), and $n=7$.

Step 2: Find ‘r’ by focusing on the variable powers.

The general term is $T_{r+1} = \binom{7}{r} (y)^{7-r} (-5)^r$.

We need the term with $y^4$. Looking at the ‘a’ part, $(y)^{7-r}$, we need the power to be 4. So, we set $7-r = 4$.

Solving for ‘r’ gives us $r=3$.

Step 3: Substitute and solve for the coefficient.

Now we plug $r=3$ into the formula, but we only care about the numerical parts.

The full term is $T_{3+1} = \binom{7}{3} (y)^{7-3} (-5)^3$.

The coefficient is just the number part: $\binom{7}{3} (-5)^3$.

$\binom{7}{3} = 35$.

$(-5)^3 = -125$.

Coefficient = $35 \times (-125) = -4375$.

Answer: The coefficient of the $y^4$ term is $-4375$.

Finding the Term Independent of a Variable

This sounds fancy, but ‘independent of x’ just means the ‘constant term’—the one where the power of x is zero (since $x^0 = 1$). This is crucial for solving the first part of our farm mystery!

Pro Tip: Rewrite Your Terms!

When you see fractions like $\frac{2}{x}$ or $\frac{1}{x^2}$, your first step should ALWAYS be to rewrite them using negative exponents. This makes applying the exponent rules way easier!

  • $\frac{2}{x}$ becomes $2x^{-1}$
  • $\frac{1}{x^2}$ becomes $x^{-2}$

Example: Finding the constant term

Problem: Find the term independent of $x$ in the expansion of $(x^3 + \frac{2}{x})^8$.

Step 1: Rewrite and identify components.

First, rewrite the expression: $(x^3 + 2x^{-1})^8$.

Now, identify: $a = x^3$, $b = 2x^{-1}$, and $n=8$.

Step 2: Set up an equation for the powers of x and solve for ‘r’.

The general term is $T_{r+1} = \binom{8}{r} (x^3)^{8-r} (2x^{-1})^r$.

Let’s just look at the ‘x’ parts: $(x^3)^{8-r} (x^{-1})^r$.

Using exponent rules, this becomes $x^{3(8-r)} \cdot x^{-r} = x^{24-3r} \cdot x^{-r} = x^{24-4r}$.

For the term to be independent of x, this final exponent must be 0. So, we set up the equation: $24 – 4r = 0$.

Solving for ‘r’, we get $4r = 24$, which means $r=6$.

Step 3: Substitute ‘r’ back in to find the constant term.

Now we calculate the full term for $r=6$:

$T_{6+1} = T_7 = \binom{8}{6} (x^3)^{8-6} (2x^{-1})^6$

$T_7 = \binom{8}{6} (x^3)^2 (2)^6 (x^{-1})^6$

$T_7 = \binom{8}{6} x^6 \cdot 64 \cdot x^{-6}$

As expected, the $x^6$ and $x^{-6}$ cancel out! We are left with the constant:

$T_7 = \binom{8}{6} \cdot 64 = 28 \cdot 64 = 1792$.

Answer: The term independent of x is 1792.

An HL Challenge: Coefficients with a Twist

For the AA HL students out there, problems often involve coefficients within the ‘a’ or ‘b’ terms themselves, which requires one extra layer of careful calculation.

IB HL Style Question

Problem: Find the coefficient of $x^5$ in the expansion of $(3x^2 – \frac{1}{x})^7$.

Step 1: Rewrite and identify.

The expression is $(3x^2 – x^{-1})^7$.

$a = 3x^2$, $b = -x^{-1}$, and $n=7$.

Step 2: Set up the exponent equation for x.

The general term’s variable part comes from $(a)^{n-r}(b)^r$, which is $(3x^2)^{7-r}(-x^{-1})^r$.

Let’s isolate the numbers and the ‘x’ parts:

$(3)^{7-r} (x^2)^{7-r} \cdot (-1)^r (x^{-1})^r$

Combining the ‘x’ powers: $x^{2(7-r)} \cdot x^{-r} = x^{14-2r-r} = x^{14-3r}$.

We need the term with $x^5$, so we set the exponent equal to 5: $14 – 3r = 5$.

Solving for ‘r’: $3r = 9$, which means $r=3$.

Step 3: Calculate the coefficient using r=3.

The coefficient is made up of all the non-x parts of the general term: $\binom{n}{r} (3)^{n-r} (-1)^r$.

Substitute $n=7$ and $r=3$:

Coefficient = $\binom{7}{3} (3)^{7-3} (-1)^3$

Coefficient = $35 \cdot (3)^4 \cdot (-1)$

Coefficient = $35 \cdot 81 \cdot (-1) = -2835$.

Answer: The coefficient of the $x^5$ term is $-2835$.

Solving the Vertical Farm Mystery

Okay, team, it’s time. We have the skills, now let’s apply them to the original problem and help out those botanists!

The model is $(x – 3y)^7$, so we have: $a=x$, $b=-3y$, and $n=7$.

Mystery #1: The ‘Stable Yield’ (Term independent of x)

We need the term where the power of $x$ is 0. The general term’s $x$ component comes from $a^{n-r}$, which is $x^{7-r}$.

We set the exponent to 0: $7-r = 0$, which gives us $r=7$.

Now we calculate the full term for $r=7$:

$T_{7+1} = T_8 = \binom{7}{7} (x)^{7-7} (-3y)^7$

$T_8 = 1 \cdot x^0 \cdot (-3)^7 y^7$

$T_8 = 1 \cdot 1 \cdot (-2187)y^7 = -2187y^7$.

Solution 1: The stable yield value, which is independent of light intensity (x), is $-2187y^7$. This tells the botanists that even with no variation in light, the yield is still heavily dependent on the nutrient density (y).

Mystery #2: The ‘Peak Performance’ Coefficient (of the $x^3$ term)

We need the coefficient of the term where the power of $x$ is 3. Again, we use the exponent from $x^{7-r}$.

We set the exponent to 3: $7-r = 3$, which gives us $r=4$.

Now we find the term for $r=4$:

$T_{4+1} = T_5 = \binom{7}{4} (x)^{7-4} (-3y)^4$

$T_5 = 35 \cdot x^3 \cdot (-3)^4 y^4$

$T_5 = 35 \cdot x^3 \cdot 81y^4$

$T_5 = 2835x^3y^4$.

The question asks for the ‘impact factor’, or the coefficient. The coefficient is everything that isn’t $x^3$.

Solution 2: The coefficient of the $x^3$ configuration is $2835y^4$. This is the precise numerical multiplier for the peak performance mode.

Mystery solved! By methodically applying the general term formula, we’ve decoded the system.

Key Takeaways & Your Turn to Practice

The Binomial Theorem might look intimidating, but as you’ve seen, most IB problems boil down to mastering the General Term Formula: $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.

Your key skills are:

  1. Identifying $a$, $b$, and $n$ correctly (especially the signs!).
  2. Setting up and solving an equation for the powers of a variable to find ‘r’.
  3. Carefully substituting ‘r’ back into the formula to find either the whole term or just the coefficient.

Remember, this isn’t just an algebra topic; it’s the foundation for understanding binomial probability later on. So, getting comfortable with it now is a huge win!

Have questions or want to discuss a problem? Share your thoughts in the comments below! Engaging with the material and your peers is a fantastic way to deepen your understanding and analytical skills in mathematics.

Additional Practice

Question 1: Find the 3rd term in the expansion of $(x + 4y)^5$.

Show Answer

Solution:

Identify: $a=x$, $b=4y$, $n=5$.

For the 3rd term, $r+1=3$, so $r=2$.

$T_3 = \binom{5}{2} x^{5-2} (4y)^2 = 10 \cdot x^3 \cdot 16y^2 = 160x^3y^2$.

Question 2: Find the 6th term in the expansion of $(2a – b)^7$.

Show Answer

Solution:

Identify: $a=2a$, $b=-b$, $n=7$.

For the 6th term, $r+1=6$, so $r=5$.

$T_6 = \binom{7}{5} (2a)^{7-5} (-b)^5 = 21 \cdot (2a)^2 \cdot (-b^5) = 21 \cdot 4a^2 \cdot (-b^5) = -84a^2b^5$.

Question 3: Find the coefficient of $x^5$ in the expansion of $(x+3)^9$.

Show Answer

Solution:

Identify: $a=x$, $b=3$, $n=9$.

The power of $x$ comes from $a^{n-r} = x^{9-r}$. We need $9-r=5$, so $r=4$.

The coefficient is $\binom{9}{4} b^4 = \binom{9}{4} (3)^4 = 126 \cdot 81 = 10206$.

Question 4: Find the coefficient of the $a^2b^4$ term in the expansion of $(a-2b)^6$.

Show Answer

Solution:

Identify: $A=a$, $B=-2b$, $n=6$.

The power of $B$ is $r$. We need the term with $b^4$, so $r=4$.

The term is $T_5 = \binom{6}{4} (a)^{6-4} (-2b)^4 = 15 \cdot a^2 \cdot (16b^4) = 240a^2b^4$.

The coefficient is 240.

Question 5: Find the term independent of $x$ in the expansion of $(x – \frac{5}{x})^4$.

Show Answer

Solution:

Rewrite: $(x – 5x^{-1})^4$. Identify $a=x$, $b=-5x^{-1}$, $n=4$.

The power of $x$ is $x^{4-r} \cdot (x^{-1})^r = x^{4-r-r} = x^{4-2r}$.

Set power to 0: $4-2r=0 \implies r=2$.

The term is $T_3 = \binom{4}{2} (x)^{4-2} (-5x^{-1})^2 = 6 \cdot x^2 \cdot 25x^{-2} = 150$.

Question 6: Find the constant term in the expansion of $(2x^2 + \frac{1}{x})^6$.

Show Answer

Solution:

Rewrite: $(2x^2 + x^{-1})^6$. Identify $a=2x^2$, $b=x^{-1}$, $n=6$.

The power of $x$ is $(x^2)^{6-r} \cdot (x^{-1})^r = x^{12-2r-r} = x^{12-3r}$.

Set power to 0: $12-3r=0 \implies r=4$.

The term is $T_5 = \binom{6}{4} (2x^2)^{6-4} (x^{-1})^4 = 15 \cdot (2x^2)^2 \cdot x^{-4} = 15 \cdot 4x^4 \cdot x^{-4} = 60$.

Question 7: Find the middle term in the expansion of $(3x – \frac{y}{3})^{8}$.

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Solution:

An expansion to the power of 8 has $8+1=9$ terms. The middle term is the 5th term.

For the 5th term, $r=4$. Identify $a=3x$, $b=-\frac{y}{3}$, $n=8$.

$T_5 = \binom{8}{4} (3x)^{8-4} (- rac{y}{3})^4 = 70 \cdot (3x)^4 \cdot (\frac{y^4}{3^4}) = 70 \cdot (81x^4) \cdot (\frac{y^4}{81})$.

The $81$s cancel, leaving $70x^4y^4$.

Question 8: The third term in the expansion of $(x+k)^5$ is $40x^3$. Find the value of $k$.

Show Answer

Solution:

Identify $a=x$, $b=k$, $n=5$. For the third term, $r=2$.

$T_3 = \binom{5}{2} x^{5-2} k^2 = 10 x^3 k^2$.

We are given that this term is $40x^3$.

So, $10x^3k^2 = 40x^3$. Divide both sides by $10x^3$: $k^2 = 4$.

Therefore, $k = \pm 2$.

Question 9: Find the coefficient of $x$ in the expansion of $(x^2 – \frac{2}{x^3})^3$.

Show Answer

Solution:

Rewrite: $(x^2 – 2x^{-3})^3$. Identify $a=x^2$, $b=-2x^{-3}$, $n=3$.

The power of $x$ is $(x^2)^{3-r} \cdot (x^{-3})^r = x^{6-2r-3r} = x^{6-5r}$.

Set power to 1 (for $x^1$): $6-5r=1 \implies 5r=5 \implies r=1$.

The coefficient comes from $\binom{3}{1} (-2)^1 = 3 \cdot (-2) = -6$.

Question 10 (IB Exam Style Question): In the expansion of $(ax + \frac{1}{x})^6$, the constant term is 160. Given that $a > 0$, find the value of $a$.

Show Answer

Solution:

Identify: $A=ax$, $B=x^{-1}$, $n=6$.

First, find the value of ‘r’ that gives the constant term.

The power of $x$ is $(x)^{6-r} \cdot (x^{-1})^r = x^{6-r-r} = x^{6-2r}$.

For the constant term, set the power to 0: $6-2r=0 \implies r=3$.

Now, write out the full constant term using $r=3$.

$T_4 = \binom{6}{3} (ax)^{6-3} (x^{-1})^3 = 20 \cdot (ax)^3 \cdot x^{-3} = 20 \cdot a^3 x^3 \cdot x^{-3} = 20a^3$.

We are told this constant term is equal to 160.

$20a^3 = 160$.

$a^3 = \frac{160}{20} = 8$.

$a = \sqrt[3]{8} = 2$.

Since we are given $a>0$, our final answer is $a=2$.

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