# Binomial Theorem

## What is a Binomial Theorem?

The expansion of the equation will become longer as the power rises and it will be very mind-numbing to calculate. Here comes the solution; a binomial expression has been improved to solve a very large power with ease by using the binomial theorem.

Let’s study all the facts associated with binomial theorem such as its definition, properties, examples, applications, etc. It will clarify all your doubts regarding the binomial theorem.

We can explain a binomial theorem as the technique to expand an expression which has been elevated to any finite power. It is a powerful tool for the expansion of the equation which has a vast use in Algebra, probability, etc.

### Binomial Theorem Expansion

In binomial theorem expansion, the binomial expression is most important in an algebraic equation which holds two different terms.

Such as: a + b, a3 + b3, etc.

Let’s consider; x, y ∈ R; n ∈ N

Then the result will be

$\sum_{i=0}^{n}nC_rx^{n-r}.y^r + nC_rx^{n-r}.y^r + ...nC_{n-1}x.y^{n-1} + nC_n.y^n$

i.e. $(x+y)^n$ = $\sum_{i=0}^{n}nC_rx^{n-r}.y^r$

Where $nC_r$ = $\frac{n!}{(n-r)!r!}$

### Binomial Theorem Examples

Example 1:

Expand the given equation (x/2 + 3/y)4.

Sol: Using the equation, $(x+y)^n$ = $\sum_{i=0}^{n}nC_rx^{n-r}.y^r$ , we get:

$\frac{x}{2} + \frac{3}{y})^4 = 4C_0(\frac{x}{2})^4 + 4C_1(\frac{x}{2})^3(\frac{3}{y}) + 4C_2(\frac{x}{2})^2(\frac{3}{y})^2+ 4C_3(\frac{x}{2})(\frac{3}{y})^3 + 4C_4(\frac{3}{y})^4$

Example 2: Expand the given equation (√3 + 1)5 + (√3 − 1)5

Sol:

We have

$(x+y)^5+(x-y)^5$ = $2(5C_0x^5 + 5C_2x^3y^2 + 5C_4xy^4)$

= $2(x^5 + 10x^3y^2 + 5xy^4)$

After applying it  to the equation now,

=  $(\sqrt{3}+ 1)^5 - (\sqrt{3}- 1)^5$

= 2[ $(\sqrt{3}^5)$ + 10$(\sqrt{3}^3)(1)^2$ +5$(\sqrt{3})(1)^4$ ]

= $88 \sqrt{3}$

### Significant Facts for Summing up Binomial Expansion

1. The whole amount of terms in the expansion of (x + y)n are (n + 1).

2. The summation of exponents of x and y is always n.

3. Binomial coefficients are known as nC0, nC1, nC2,…up to n Cn, and similarly signified by C0, C1, C2, ….., Cn.

4. The binomial coefficients which are intermediate from the start and the finish are equal i.e. nC= nCn, nC= nCn-1, nC= nCn-2,….. etc.

Also, we can apply Pascal’s triangle to find binomial coefficients.

Also, there are some other important expansions that you must learn about. They are given below and many useful binomial expansions.

1. (1+x)+ (1 − x)= 2[K0 + K2 x2 + K4 x+ …]

2. (x + y)– (x−y)= 2[K1 xn-1 y + K3 xn-3 y+ K5 xn-5 y+ …]

3. (1 + x)n nΣr-0 n C. x = [K+ K1 x + K2 x+ … Kn xn]

4. (x + y)+ (x−y)= 2[K0 x+ K2 xn-1 y+ K4 xn-4 y+ …]

5. (1+x)− (1−x)= 2[K1 x + K3 x3 + K5 x5 + …]

6. In the expansion (x + a)n − (x−a); the number of terms of are (n/2) if “n” is even or (n+1)/2 if “n” is odd.

7. In the expansion (x + a)n + (x−a)n; the number of terms are (n+2)/2 if “n” is even or (n+1)/2 if “n” is odd.

### Binomial Coefficient Properties

Here, I am considering the binomial coefficient as K. Some of the most important properties of binomial coefficients are:

1. K+ K+ K+ … = K+ K+ K+ … = 2n-1

2. K+ K+ K+ … + K= 2n

3. nK+ 2.nK+ 3.nK+ … + n.nK= n.2n-1

4. K– K+ K– K+ … +(−1)n.nK= 0

5. K0+ K1+ K2+ …Kn= [(2n)!/ (n!)2]

6. K– 2K+ 3K– 4K+ … +(−1)n-1 K= 0 for n > 1

### Application of Binomial Theorem

1. To Calculate the value ‘e’ (Euler's Number)

As we know, e = 2.71828182846...

Here, e = (1 + 1/n)n

Now it is time to apply Binomial Theorem:

(1+1/n)n=$\sum_{k=0}^{n}(nk)1^{(n-k)}(\frac{1}{n})^k$= $\sum_{k=0}^{n}(nk)(\frac{1}{n})^k$

To obtain the most precise value of e, the amount of ‘n’ should be as large as possible.

Here, $\lim_{n\rightarrow\infty}$ is expressing that ‘n’ should be the largest possible number.

= $\lim_{n\rightarrow\infty}\sum_{K=0}^{n}\frac{n!}{k!(n-k)!}.\frac{1}{n^k}$

Here,

$\lim_{n\rightarrow\infty}\frac{n!}{k!(n-k)!}.\frac{1}{n^k}$ = $\lim_{n\rightarrow\infty}\frac{n}{n}.\frac{n-1}{n}\frac{n-2}{n}...\frac{n-k}{n}$ = 1⋅1⋅1⋅……1(as n→∞)

Now the obtained equation can be written as:

$\sum_{K=0}^{n}\frac{1}{k!}$ = $\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!}$ +......= 1 + 1 + $\frac{1}{2}$ + $\frac{1}{6}$ + $\frac{1}{24}$ + ...

By calculating the above equation for certain terms we calculate that

e = 2.7083

The final expression is obtained by the use of supercomputers to find the most perfect value of e.

1. Divisibility and remainder problems is an application of binomial theorem

Example- If 3230 is divided by 7, then what will be the remainder?

As we know, 25 = 32,

3230 can thus be written as:

(25)30 = 2150 = (23)50 = 850 = (7 + 1)50

= [ (7)50 + 50K1(7)49 + 50K2 (7)48 + ...  + 1]

= [7 ( (7)49 + 50K1(7)48  +  50K2 (7)47  + ... ) + 1 ]

Now, (7k + 1)/7, gives us the remainder 1

1. Concluding the highest term is an application of binomial theorem

• If n>6, then $(\frac{n}{3})^n$ < n! < $(\frac{n}{2})^n$

• n ≥ 1 and n ∈ N, 2 ≤ (1+1/n) < 3

The given theorem can be utilized in solving problems such as, which one is greatest among 100100 and (300)!

From the above results,

Insert n = 300

= (100)300 < (300)! .........(i)

But, (100)300 > (100)100 .........(ii)

From equation (i) and (ii)

= (100)100 < (100)300 < (300)!

Therefore, (100)100 < (300)!

Q1. Describe the usage of the Binomials.

Ans: The binomial distribution model permits us to calculate the probability of noticing a definite number of "successes" when the procedure is repetitive for a specific amount of times such as in a set of patients and the outcome for a specified patient is either a victory or a disappointment.

Ans: The binomial distribution is highly important as it has a huge range of applications. The only reason is that its heart is a binary situation: one with two likely results.

We present a binomial unintended variable as the number of 'successes' in ‘n’ known as ‘independent Bernoulli trials, each having the same probability of success ‘p’.

Q3. Can you elaborate on the binomial theorem in statistics?

Ans: The binomial theorem which is occasionally known as binomial expansion is the most common method which is used in statistics as a simple formula.

This method (formula) is applied to calculate the probabilities for binomial experiments for the events which have two choices such as heads or tails.

Q4. Let’s calculate the coefficient of y9 in the expansion of (1 + y) (1 + y2) (1 + y3) . . . . . . (1 + y100).

Ans:

Here, y9 can be fashioned in 8 ways.

Such as y9, y1+8, y2+7, y3+6,y4+5, y1+3+5, y2+3+4

∴ The coefficient of y9 = 1 + 1 + 1 + . . . . + 8 times = 8.