Binomial Theorem and Its Applications

Binomial theorem for positive integral index

The binomial theorem provides a formula for expanding expressions of the form (x + y)ⁿ, where n is a non-negative integer. This theorem is incredibly useful in algebra, calculus, and probability. We will focus here on the case where the index 'n' is a positive integer.

Understanding the Expansion

Let's look at a few simple cases to see the pattern:

• For n=1: (x + y)¹ = x + y
• For n=2: (x + y)² = x² + 2xy + y²
• For n=3: (x + y)³ = x³ + 3x²y + 3xy² + y³
• For n=4: (x + y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴

Observe the following patterns in these expansions:

• The first term is always xⁿ and the last term is always yⁿ.
• The powers of x decrease by 1 in each successive term, starting from n and going down to 0.
• The powers of y increase by 1 in each successive term, starting from 0 and going up to n.
• The sum of the powers of x and y in each term is always equal to n.
• The coefficients of the terms follow a specific pattern.

The Binomial Coefficients and Pascal's Triangle

The coefficients in the binomial expansion are called binomial coefficients. They can be represented using the combination notation C(n, k) or "n choose k", written as $\binom{n}{k}$ or ⁿC_k. This represents the number of ways to choose k items from a set of n distinct items. The formula for $\binom{n}{k}$ is:

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

where '!' denotes the factorial (e.g., 5! = 5 × 4 × 3 × 2 × 1).

Let's look at the coefficients from our examples:

• n=1: 1, 1 which are $\binom{1}{0}$, $\binom{1}{1}$
• n=2: 1, 2, 1 which are $\binom{2}{0}$, $\binom{2}{1}$, $\binom{2}{2}$
• n=3: 1, 3, 3, 1 which are $\binom{3}{0}$, $\binom{3}{1}$, $\binom{3}{2}$, $\binom{3}{3}$
• n=4: 1, 4, 6, 4, 1 which are $\binom{4}{0}$, $\binom{4}{1}$, $\binom{4}{2}$, $\binom{4}{3}$, $\binom{4}{4}$

These coefficients can be visualized using Pascal's Triangle:

Row 0: 1
Row 1: 1 1
Row 2: 1 2 1
Row 3: 1 3 3 1
Row 4: 1 4 6 4 1

Each number in Pascal's Triangle is the sum of the two numbers directly above it. The nth row (starting from row 0) gives the binomial coefficients for the expansion of (x + y)ⁿ.

The Binomial Theorem Statement

For any positive integer n, the binomial theorem states:

(x + y)ⁿ = $\binom{n}{0}$xⁿy⁰ + $\binom{n}{1}$x^n-1y¹ + $\binom{n}{2}$x^n-2y² + ... + $\binom{n}{r}$x^n-ry^r + ... + $\binom{n}{n}$x⁰yⁿ

This can be written more compactly using summation notation:

(x + y)ⁿ = $\sum_{r=0}^{n} \binom{n}{r} x^{n-r} y^{r}$

Here, 'r' is the index of summation, starting from 0 and going up to n.

The General Term

In the expansion of (x + y)ⁿ, the (r+1)^th term (often denoted as T_r+1) is given by:

T_r+1 = $\binom{n}{r} x^{n-r} y^{r}$

This is a crucial formula for finding a specific term in the expansion without calculating the entire expansion. Remember that 'r' starts from 0 for the first term. So, if you need the 5th term, you set r = 4.

Example 1: Expanding a Binomial

Expand (a + 2b)⁴.

Here, x = a, y = 2b, and n = 4.

Using the binomial theorem:

(a + 2b)⁴ = $\sum_{r=0}^{4} \binom{4}{r} a^{4-r} (2b)^{r}$

Let's calculate each term:

• r=0: $\binom{4}{0} a^{4-0} (2b)^{0} = 1 \cdot a^4 \cdot 1 = a^4$
• r=1: $\binom{4}{1} a^{4-1} (2b)^{1} = 4 \cdot a^3 \cdot 2b = 8a^3b$
• r=2: $\binom{4}{2} a^{4-2} (2b)^{2} = 6 \cdot a^2 \cdot 4b^2 = 24a^2b^2$
• r=3: $\binom{4}{3} a^{4-3} (2b)^{3} = 4 \cdot a^1 \cdot 8b^3 = 32ab^3$
• r=4: $\binom{4}{4} a^{4-4} (2b)^{4} = 1 \cdot a^0 \cdot 16b^4 = 16b^4$

So, (a + 2b)⁴ = a⁴ + 8a³b + 24a²b² + 32ab³ + 16b⁴.

Example 2: Finding a Specific Term

Find the 5th term in the expansion of (3x - 2y)⁷.

Here, x = 3x, y = -2y, and n = 7.

We need the 5th term. Using the shortcut, the 5th term corresponds to r = 5 - 1 = 4.

The general term is T_r+1 = $\binom{n}{r} x^{n-r} y^{r}$.

For the 5th term (r=4):

T₅ = T₄₊₁ = $\binom{7}{4} (3x)^{7-4} (-2y)^{4}$

Calculate the binomial coefficient:

$\binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

Now substitute back into the term formula:

T₅ = 35 (3x)³ (-2y)⁴

T₅ = 35 (27x³) (16y⁴)

T₅ = 35 × 27 × 16 × x³y⁴

T₅ = 15120 x³y⁴

Properties of Binomial Coefficients

The binomial coefficients $\binom{n}{r}$ have several important properties:

• Symmetry: $\binom{n}{r} = \binom{n}{n-r}$. This means the coefficients are symmetric around the middle of the expansion. For example, in (x+y)⁴, the coefficients are 1, 4, 6, 4, 1. $\binom{4}{1} = 4$ and $\binom{4}{3} = 4$.
• Sum of Coefficients: The sum of the binomial coefficients for a given n is 2ⁿ. This can be seen by setting x=1 and y=1 in the binomial expansion: (1 + 1)ⁿ = $\sum_{r=0}^{n} \binom{n}{r} (1)^{n-r} (1)^{r} = \sum_{r=0}^{n} \binom{n}{r}$. So, $\sum_{r=0}^{n} \binom{n}{r} = 2^n$.
• Alternating Sum: The alternating sum of binomial coefficients is 0. This is found by setting x=1 and y=-1: (1 - 1)ⁿ = $\sum_{r=0}^{n} \binom{n}{r} (1)^{n-r} (-1)^{r} = \sum_{r=0}^{n} \binom{n}{r} (-1)^{r}$. So, $\sum_{r=0}^{n} \binom{n}{r} (-1)^{r} = 0$ (for n > 0).

Example 3: Using Properties of Coefficients

Find the sum of coefficients in the expansion of (2x - 3y)⁵.

To find the sum of coefficients, set x=1 and y=1.

Sum = (2(1) - 3(1))⁵ = (2 - 3)⁵ = (-1)⁵ = -1.

Find the sum of the coefficients of the odd power terms of x in the expansion of (x + y)ⁿ.

Let the expansion be E = $\sum_{r=0}^{n} \binom{n}{r} x^{n-r} y^{r}$.

Let E₁ be the expansion with all positive terms (x=1, y=1): E₁ = (1+1)ⁿ = 2ⁿ = $\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \binom{n}{3} + ...$

Let E₂ be the expansion with alternating terms (x=1, y=-1): E₂ = (1-1)ⁿ = 0 (for n>0) = $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + ...$

Subtracting E₂ from E₁:

E₁ - E₂ = 2ⁿ - 0 = 2ⁿ

( $\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \binom{n}{3} + ...$) - ( $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + ...$)

= $2\binom{n}{1} + 2\binom{n}{3} + 2\binom{n}{5} + ...$

So, $2^n = 2(\binom{n}{1} + \binom{n}{3} + \binom{n}{5} + ...)$

Therefore, the sum of coefficients of odd power terms of x (which correspond to odd values of r) is $\binom{n}{1} + \binom{n}{3} + \binom{n}{5} + ... = 2^{n-1}$.

Example 4: Middle Term(s)

Find the middle term in the expansion of (x + y)¹⁰.

Since n = 10 (an even number), there will be n+1 = 11 terms in the expansion. The middle term is the (11+1)/2 = 6th term.

The 6th term means r+1 = 6, so r = 5.

Using the general term formula T_r+1 = $\binom{n}{r} x^{n-r} y^{r}$:

T₆ = $\binom{10}{5} x^{10-5} y^{5}$

Calculate the coefficient:

$\binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 2 \times 9 \times 2 \times 7 = 252$

So, the middle term is 252x⁵y⁵.

Find the middle terms in the expansion of (x - 2y)⁹.

Since n = 9 (an odd number), there will be n+1 = 10 terms. There are two middle terms: the 10/2 = 5th term and the (10/2) + 1 = 6th term.

For the 5th term: r+1 = 5, so r = 4.

T₅ = $\binom{9}{4} (x)^{9-4} (-2y)^{4}$

$\binom{9}{4} = \frac{9!}{4!5!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 9 \times 2 \times 7 = 126$

T₅ = 126 (x)⁵ (16y⁴) = 126 × 16 x⁵y⁴ = 2016 x⁵y⁴.

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

T₆ = $\binom{9}{5} (x)^{9-5} (-2y)^{5}$

Note that $\binom{9}{5} = \binom{9}{9-5} = \binom{9}{4} = 126$.

T₆ = 126 (x)⁴ (-32y⁵) = 126 × (-32) x⁴y⁵ = -4032 x⁴y⁵.

The middle terms are 2016 x⁵y⁴ and -4032 x⁴y⁵.

Special Cases of the Binomial Theorem

1. Expansion of (1 + x)ⁿ:

If we substitute 'a' for 'x' and '1' for 'y' in the general binomial theorem, we get (1 + x)ⁿ. The formula becomes:

(1 + x)ⁿ = $\binom{n}{0}(1)^{n}x^{0} + \binom{n}{1}(1)^{n-1}x^{1} + \binom{n}{2}(1)^{n-2}x^{2} + ... + \binom{n}{n}(1)^{0}x^{n}$

(1 + x)ⁿ = $\binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^{2} + ... + \binom{n}{n}x^{n}$

Or, $\sum_{r=0}^{n} \binom{n}{r} x^{r}$. This form is very common, especially when dealing with probabilities and series.

2. Expansion of (1 - x)ⁿ:

This is obtained by replacing x with -x in the expansion of (1 + x)ⁿ:

(1 - x)ⁿ = $\binom{n}{0} + \binom{n}{1}(-x) + \binom{n}{2}(-x)^{2} + \binom{n}{3}(-x)^{3} + ... + \binom{n}{n}(-x)^{n}$

(1 - x)ⁿ = $\binom{n}{0} - \binom{n}{1}x + \binom{n}{2}x^{2} - \binom{n}{3}x^{3} + ... + (-1)^{n}\binom{n}{n}x^{n}$

Or, $\sum_{r=0}^{n} \binom{n}{r} (-x)^{r} = \sum_{r=0}^{n} \binom{n}{r} (-1)^{r} x^{r}$.

Example 5: Using (1+x)ⁿ Expansion

Find the coefficient of x³ in the expansion of (1 + 2x)⁶.

This is in the form (1 + X)ⁿ where X = 2x and n = 6.

The general term in the expansion of (1 + X)ⁿ is $\binom{n}{r} X^{r}$.

Here, the general term is $\binom{6}{r} (2x)^{r} = \binom{6}{r} 2^{r} x^{r}$.

We want the term with x³, so we set r = 3.

The term is $\binom{6}{3} 2^{3} x^{3}$.

Calculate the coefficient:

$\binom{6}{3} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

The coefficient is $20 \times 2^{3} = 20 \times 8 = 160$.

So, the term is 160x³, and the coefficient of x³ is 160.

Example 6: Using (1-x)ⁿ Expansion

Find the coefficient of x⁴ in the expansion of (3 - x)⁵.

First, rewrite the expression in the form of (1 - X)ⁿ or (a+bx)ⁿ.

(3 - x)⁵ = [3(1 - x/3)]⁵ = 3⁵ (1 - x/3)⁵.

Now, expand (1 - x/3)⁵ using the formula for (1 - X)ⁿ, where X = x/3 and n = 5.

(1 - x/3)⁵ = $\sum_{r=0}^{5} \binom{5}{r} (-x/3)^{r} = \sum_{r=0}^{5} \binom{5}{r} (-1)^{r} (x/3)^{r}$

We want the term with x⁴, which means r = 4.

The term with x⁴ in the expansion of (1 - x/3)⁵ is:

$\binom{5}{4} (-1)^{4} (x/3)^{4} = 5 \cdot 1 \cdot \frac{x^{4}}{3^{4}} = 5 \frac{x^{4}}{81}$.

Now, multiply by the 3⁵ factor we took out earlier:

3⁵ × (5/81) x⁴ = 243 × (5/81) x⁴.

Simplify: 243/81 = 3.

So, the term is 3 × 5 x⁴ = 15x⁴.

The coefficient of x⁴ is 15.

Binomial Expansion of (x+y)ⁿ vs (x-y)ⁿ

It's important to distinguish between (x+y)ⁿ and (x-y)ⁿ.

(x + y)ⁿ = $\sum_{r=0}^{n} \binom{n}{r} x^{n-r} y^{r}$ (all terms positive if x and y are positive)

(x - y)ⁿ = $\sum_{r=0}^{n} \binom{n}{r} x^{n-r} (-y)^{r}$ (terms alternate in sign: +, -, +, -, ...)

(x - y)ⁿ = $\binom{n}{0}x^{n} - \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^{2} - \binom{n}{3}x^{n-3}y^{3} + ... + (-1)^{n}\binom{n}{n}y^{n}$

When using the general term formula T_r+1 = $\binom{n}{r} x^{n-r} y^{r}$, if the binomial is (x-y)ⁿ, you must include the negative sign with 'y' when substituting.

Example 7: Handling Negative Terms

Find the 4th term in the expansion of (x - 2)⁵.

Here, x is 'x', y is '2', and the sign is negative, so n = 5.

We need the 4th term, so r+1 = 4, which means r = 3.

T_r+1 = $\binom{n}{r} x^{n-r} y^{r}$

T₄ = $\binom{5}{3} (x)^{5-3} (-2)^{3}$

$\binom{5}{3} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$.

T₄ = $10 \cdot x^{2} \cdot (-8)$

T₄ = -80x².

The 4th term is -80x².

Summary of Key Formulas and Concepts

For a positive integer n:

• Binomial Expansion: $(x + y)^n = \sum_{r=0}^{n} \binom{n}{r} x^{n-r} y^{r}$
• General Term: $T_{r+1} = \binom{n}{r} x^{n-r} y^{r}$
• Binomial Coefficient: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$
• Expansion of (1 + x)ⁿ: $\sum_{r=0}^{n} \binom{n}{r} x^{r} = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^{2} + ... + \binom{n}{n}x^{n}$
• Expansion of (1 - x)ⁿ: $\sum_{r=0}^{n} \binom{n}{r} (-1)^{r} x^{r} = \binom{n}{0} - \binom{n}{1}x + \binom{n}{2}x^{2} - ... + (-1)^{n}\binom{n}{n}x^{n}$
• Sum of Coefficients: Set x=1, y=1 in (x+y)ⁿ, sum = 2ⁿ.
• Middle Term(s): If n is even, the middle term is the (n/2 + 1)^th term. If n is odd, the middle terms are the (n+1)/2^th and (n+3)/2^th terms.