Last modified on February 26th, 2024

chapter outline

 

Binomial Theorem

The binomial theorem is a formula for expanding binomial expressions of the form (x + y)n, where ‘x’ and ‘y’ are real numbers and n is a positive integer.

The simplest binomial expression x + y with two unlike terms, ‘x’ and ‘y’, has its exponent 0, which gives a value of 1

(x + y)0  = 1

If the exponent is increased by 1, (x + y)1  it gives the value x + y

(x + y)1 = x + y

Increasing the exponent further by 1, the binomial expression becomes (x + y)2, which can be written as (x + y)(x + y)

On multiplying the binomials and using the distributive property, we get x(x + y) + y(x + y)

⇒ (x + y)2 = x2 + 2xy + y2

The result can be geometrically represented in 2 dimensions as:

Binomial Square in Geometry

Now, if we multiply the result with (x + y) again, we get (x2 + 2xy + y2) (x + y)

⇒ x(x2 + 2xy + y2) + y(x2 + 2xy + y2)

⇒ (x + y)3 = x3 + 3x2y + 3xy2 + y3

The result can be geometrically represented in 3 dimensions as:

Binomial Cube in Geometry

The calculation gets longer and more complicated with each round of multiplication. However, we can observe a specific pattern developing in the results, which can be summed up using the binomial formula:

Binomial Theorem Formula

Using the above formula, we can summarize the expansions till (x + y)n. Here, we show the expansions up to (x + y)8

The Pattern

Binomial Theorem

Let us go back to the expression (x2 + 2xy + y2). Here, the exponents of x start from 2 and go down to 1, 0.

Similarly, in the expression (x3 + 3x2y + 3xy2 + y3), the exponents of x start from 3 and go down as 2, 1, 0.

Now, in the expression (x2 + 2xy + y2), the exponents of y start from 0 and go up to 3. Similarly, in the expression (x3 + 3x2y + 3xy2 + y3), the exponent goes on as 0, 1, 2, 3. 

This surely follows a pattern, which, when put together, can be written as:

xn-kyk

If we replace the value of n and k in the expansion of (x + y)3, we get the terms as shown:

k = 0k = 1k = 2k = 3
xn-kyk = x3-0y0 = x3xn-kyk = x3-1y1 = x2yxn-kyk = x3-2y2 = xy2xn-kyk = x3-3y3 = y3

Thus, we conclude that any expression with the exponents of n has n + 1 number of terms in its expansion.

Relation with Pascal’s Triangle

Now, if we place all the results of the binomial expressions (x + y) with their exponents ranging from 0 till 3 and so on in the form of a triangle, we get:

Now, if we focus only on the coefficients of each term, it is found to form Pascal’s triangle.

Binomial Theorem and Pascal’s Triangle

Now, let us expand the expression (x + y)5 using Pascal’’s triangle.

Since the coefficients of the 5th row are 1, 5, 10, 10, 5, 1, the expression (x + y)5 on expanding is x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5

This is true for all exponents of the binomial expression (x + y)

The coefficients of the terms also form Pascal’s triangle when written using the combination formula. 

Combinations in Pascal’s Triangle

For example, on expanding (x + y)5, we get

${\begin{pmatrix} 5 \\ 0 \end{pmatrix}x^{5}+\begin{pmatrix} 5 \\ 1 \end{pmatrix}x^{4}y+\begin{pmatrix} 5 \\ 2 \end{pmatrix}x^{3}y^{2}+\begin{pmatrix} 5 \\ 3 \end{pmatrix}x^{2}y^{3}+\begin{pmatrix} 5 \\ 4 \end{pmatrix}xy^{4}+\begin{pmatrix} 5 \\ 5 \end{pmatrix}y^{5}}$

= x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5

Proof

We can prove the binomial theorem for all the natural numbers using the principle of mathematical induction.

Here, x, y, n, and k belong to natural numbers.

For n = 1, (x + y)1 = x + y

For n = 2, (x + y)2 = (x + y)(x + y)

Now, using the distributive property, we get (x + y)2 = x2 + 2xy + y2

Thus, the results are true for n = 1 and n = 2.

Let r (≥ 2) be any natural number, and consider n = r in (x + y)n = ${\sum ^{n}_{k=0}\begin{pmatrix} n \\ k \end{pmatrix}x^{n-k}y^{k}}$.

We get, (x + y)r = ${\sum ^{r}_{k=0}\begin{pmatrix} r \\ k \end{pmatrix}x^{r-k}y^{k}}$

⇒ (x + y)r = rC0 xry0 + rC1 xr – 1y1 + rC2 xr – 2 y2 + … + rCk xr – kyk +….+ rCr x0yr

⇒ (x + y)r = xr + rC1 xr – 1y1 + rC2 xr – 2 y2 + … + rCk xr – kyk +….+ yr

Thus, the result is true for r ≥ 2.

Now, let us consider the expansion for n = r + 1.

(x + y)r + 1 = (x + y)(x + y)r

⇒ (x + y)r + 1 = (x + y)(xr + rC1 xr – 1y1 + rC2 xr – 2 y2 + … + rCk xr – kyk +….+ yr)

⇒ (x + y)r + 1 = xr + 1 + (1 + rC1)xry + (rC1 + rC2)xr – 1y2 + … + (rCk – 1 + rCk)xr – k + 1yk + … + (rCk-1 + 1)xyr + yr+1

As we know, nCr + nCr – 1 = n + 1Cr

=> (x + y)r + 1 = xr+1 + r+1C1 xry + r+1C2 xr-1y2 + … + r+1Ck xr-k+1yk + … + r+1Cr xyr + yr+1, which is true for n = r + 1

Thus, we conclude that the result is true for all positive integers n.

Binomial Expansion for Negative Exponents

The binomial theorem also applies to exponents with negative terms. 

The terms and coefficient values remain the same, but their algebraic relation changes. 

For example, 

(1 + x)-1 = 1 – x + x2 – x3 + x4 + … 

(1 + x)-2 = 1 – 2x + 3x2 – 4x3 + 5x4 + … 

(1 + x)-3 = 1 – 3x + 6x2 – 10x3 + 15x4 + … 

Solved Examples

Expand the binomial (n2 + p)5 using the binomial theorem.

Solution:

As we know, (x + y)r = xr + rC1 xr – 1y1 + rC2 xr – 2 y2 + … + rCk xr – kyk +….+ yr
Here, (n2 + p)5 = (n2)5 + 5C1 (n2)5 – 1p1 + 5C2 (n2)5 – 2 p2 + 5C3 (n2)5 – 3p3 + 5C4 (n2)5 – 4p4 + p5
= n10 + 5n8p + 10n6p2 + 10n4p3 + 5n2p4 + p5

Find the coefficient of x3 in the expansion of (2x + 5)4.

Solution:

As we know, any kth term in the expansion of (x + y)n is written as nCk xn – kyk
Here, we want the term x3, and n = 4, that means n – k = 3 ⇒ 4 – k = 3 ⇒ k = 1
Thus, the term with x3 is 4C1 (2x)4 – 1(5)1 = 4(8x3)(5) = 160x3
The coefficient of x3 is 160.

Last modified on February 26th, 2024

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