The Binomial Theorem is a fast method of expanding or multiplying out a binomial expression. The expression has been raised to some large power. As we know the multiplication of such expressions is always troublesome with large powers and terms. But Binomial expansions and formula help a lot in this regard. In this article, we will discuss the Binomial theorem and the Binomial Theorem Formula. Let us begin the learning of the concept with examples.

**Binomial Theorem Formula**

**What is Binomial Expansion?**

The binomial theorem is used to describe the expansion in algebra for the powers of a binomial. According to this theorem, it is possible to expand the polynomial \((x + y)^n\) into a series of the sum involving terms of the form a \(x^b y^c\)

Here the exponents b and c are non-negative integers with condition that b + c = n. Also, the coefficient of each term is a specific positive integer depending on n and b.

For example (for n = 4), we have:

\((x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4\)

It is obvious that such expressions and their expansions would be very painful to multiply out by hand. Thankfully, someone has figured out a formula for this expansion, and we can easily use it.

**Binomial Theorem Formula**

The expression of the Binomial Theorem formula is given as follows:

**\((x+y)^n\)=\(\sum_{k=0}^{n}\) \({n \choose k} x^{n – k} y^k \)**

Also, Recall that the factorial notation n! Here, it represents the product of all the whole numbers between 1 and n.

Some expansions are as follows:

- \((x+y)^1 = x+y\)
- \((x+y)^2 = x^2 + 2xy +y^2\)
- \((x+y)^3 = x^3 + 3x^2y+ 3xy^2+ y^3\)

**Properties of the Binomial Expansion \((x + y)^ n\) **

- There are n+1 terms in total.
- The first term is \(x^n\) and the final term is\( y^n\)
- Progressing from the first term to the last, the exponent of x decreases by 1 from term to term. While the exponent of y increases by 1. Also, the sum of both the exponents in each term will be n.
- If the coefficient of each term is multiplied by the exponent of x in that term, and the product is divided by the number of that term, we can easily obtain the coefficient of the next term.

**Binomial Series**

From the binomial theorem formula, if we assume

x = 1 and y = x, we can also obtain the binomial series which is valid for any real number n if |x| < 1.

Thus

**(\(1+x)^n = 1 + nx + \frac{n(n-1)}{2!} x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + …….\)**

## Solved Examples

Example-1: (1) Using the binomial series, find the first four terms of the expansion: \(\sqrt[3]{1+x}\)

(2) Use your result from part (a) to approximate the value of \(\sqrt[3]{1.2}\)

Solution:

First, we will write the expansion formula for \((1+x)^3\) as follows:

\((1+x)^n = 1 + nx + \frac{n(n-1)}{2!} x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + …….\)

Put value of n =\frac{1}{3}, till first four terms:

\((1+x)^\frac{1}{3}= 1+ \frac {1}{3}x+\frac{\frac{1}{3}(\frac{1}{3}-1)}{2!} x^2 + \frac{\frac{1}{3}(\frac{1}{3}-1)(\frac{1}{3}-2)}{3!}x^3\)

Thus expansion is:

**\((1+x)^\frac{1}{3} = 1+ \frac {1}{3}x – \frac{x^2}{9}+ \frac {5x^3}{81}\)**

(2) Now put x=0.2 in above expansion to get value of \(\sqrt[3]{1.2}\)

So, \(\sqrt[3]{1.2}\)

\((1+0.2)^\frac{1}{3} = 1+ \frac {1}{3} (0.2) -\frac{(0.2)^2}{9}+ \frac {5(0.2)^3}{81}\\\)

\((1.2)^\frac{1}{3} = \frac{81+5.4-3.6+4}{81}\\\)

\((1.2)^\frac{1}{3} = 1.071\)

**Thus value is 1.071.**

I get a different answer for first example.

I got Q1 as 20.5

median 23 and

Q3 26

Hi

Same

yes