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Last modified on February 26th, 2024

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} = x^{2} + 2xy + y^{2}

The result can be geometrically represented in 2 dimensions as:

Now, if we multiply the result with (x + y) again, we get (x^{2} + 2xy + y^{2}) (x + y)

⇒ x(x^{2} + 2xy + y^{2}) + y(x^{2} + 2xy + y^{2})

⇒ (x + y)^{3} = x^{3} + 3x^{2}y + 3xy^{2} + y^{3}

The result can be geometrically represented in 3 dimensions as:

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:

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

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

Similarly, in the expression (x^{3} + 3x^{2}y + 3xy^{2} + y^{3}), the exponents of x start from 3 and go down as 2, 1, 0.

Now, in the expression (x^{2} + 2xy + y^{2}), the exponents of y start from 0 and go up to 3. Similarly, in the expression (x^{3} + 3x^{2}y + 3xy^{2} + y^{3}), the exponent goes on as 0, 1, 2, 3.

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

x^{n-k}y^{k}

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

k = 0 | k = 1 | k = 2 | k = 3 |

x^{n-k}y^{k} = x^{3-0}y^{0} = x^{3} | x^{n-k}y^{k} = x^{3-1}y^{1} = x^{2}y | x^{n-k}y^{k} = x^{3-2}y^{2} = xy^{2} | x^{n-k}y^{k} = x^{3-3}y^{3} = y^{3} |

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

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.

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

Since the coefficients of the 5^{th} row are 1, 5, 10, 10, 5, 1, the expression (x + y)^{5} on expanding is x^{5} + 5x^{4}y + 10x^{3}y^{2} + 10x^{2}y^{3} + 5xy^{4} + y^{5}

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.

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

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

= x^{5} + 5x^{4}y + 10x^{3}y^{2} + 10x^{2}y^{3} + 5xy^{4} + y^{5}

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} = x^{2} + 2xy + y^{2}

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} = _{r}C_{0 }x^{r}y^{0} + _{r}C_{1 }x^{r – 1}y^{1} + _{r}C_{2 }x^{r – 2 }y^{2} + … + _{r}C_{k }x^{r – k}y^{k} +….+ _{r}C_{r }x^{0}y^{r}

⇒ (x + y)^{r} = x^{r }+ _{r}C_{1 }x^{r – 1}y^{1 }+ _{r}C_{2 }x^{r – 2 }y^{2 }+ … + _{r}C_{k }x^{r – k}y^{k} +….+ y^{r}

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)(x^{r }+ _{r}C_{1 }x^{r – 1}y^{1 }+ _{r}C_{2 }x^{r – 2 }y^{2 }+ … + _{r}C_{k }x^{r – k}y^{k} +….+ y^{r})

⇒ (x + y)^{r + 1} = x^{r + 1} + (1 + _{r}C_{1})x^{r}y + (_{r}C_{1} + _{r}C_{2})x^{r – 1}y^{2} + … + (_{r}C_{k – 1} + _{r}C_{k})x^{r – k + 1}y^{k} + … + (_{r}C_{k-1} + 1)xy^{r} + y^{r+1}

As we know, _{n}C_{r} + _{n}C_{r – 1} = _{n + 1}C_{r}

=> (x + y)^{r + 1} = x^{r+1} +_{ r+1}C_{1 }x^{r}y + _{r+1}C_{2 }x^{r-1}y^{2} + … +_{ r+1}C_{k }x^{r-k+1}y^{k} + … +_{ r+1}C_{r }xy^{r} + y^{r+1}, which is true for n = r + 1

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

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 + x^{2} – x^{3} + x^{4} + …

(1 + x)^{-2} = 1 – 2x + 3x^{2} – 4x^{3} + 5x^{4} + …

(1 + x)^{-3} = 1 – 3x + 6x^{2} – 10x^{3} + 15x^{4} + …

**Expand the binomial (n ^{2 }+ p)^{5} using the binomial theorem.**

Solution:

As we know, (x + y)^{r} = x^{r }+ _{r}C_{1 }x^{r – 1}y^{1 }+ _{r}C_{2 }x^{r – 2 }y^{2 }+ … + _{r}C_{k }x^{r – k}y^{k} +….+ y^{r}

Here, (n^{2 }+ p)^{5} = (n^{2})^{5} + _{5}C_{1 }(n^{2})^{5 – 1}p^{1 }+ _{5}C_{2 }(n^{2})^{5 – 2 }p^{2 }+ _{5}C_{3 }(n^{2})^{5 – 3}p^{3} + _{5}C_{4 }(n^{2})^{5 – 4}p^{4} + p^{5}

= n^{10} + 5n^{8}p + 10n^{6}p^{2} + 10n^{4}p^{3} + 5n^{2}p^{4} + p^{5}

**Find the coefficient of x ^{3} in the expansion of (2x + 5)^{4}.**

Solution:

As we know, any k^{th }term in the expansion of (x + y)^{n} is written as _{n}C_{k} x^{n – k}y^{k}

Here, we want the term x^{3}, and n = 4, that means n – k = 3 ⇒ 4 – k = 3 ⇒ k = 1

Thus, the term with x^{3} is _{4}C_{1} (2x)^{4 – 1}(5)^{1} = 4(8x^{3})(5) = 160x^{3}

The coefficient of x^{3} is 160.

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Last modified on February 26th, 2024