Corollaries of Binomial Theorem

The Binomial Theorem describes how to expand an expression raised to any finite power. A binomial theorem is a powerful expansion tool that has applications in algebra, probability, and other fields.

Assume we have an expression $\mathrm{(x\:+\:y)^n}$and we need to expand the expression, we can do this using the generalised equation of binomial theorem.

The binomial theorem defines a binomial expression for two different terms. The general equation of binomial theorem is:

$$\mathrm{(a+b)^{n}=^n{C_{r=0}}a^{n-r}b^{0}\:+\:^n{C_{r=1}}a^{n-1}b^{1\:}+\:........\:+\:^n{C_{r=n-1}}a^{1}b^{n-1}+^n{C_{r=n}}a^{0}b^{n}}$$

$$\mathrm{=n_{\sum_{r=0}}^n{C_{r}}a^{n-r}b^{r}}$$

Where we can get the value of $\mathrm{^n{C_{r}}}$ using the formula,

$$\mathrm{^n{C_{r}}=\frac{n!}{(n-r)!r!}}$$[0! is always equals to 1]

NOTE

• There are total n+1 terms in the binomial expansion of the expression $\mathrm{(a\:+\:b)^{n}}$.

• The sum of power of a and b in every term will be equal to n.

For example, if we want to expand $\mathrm{(x \:+\: y)^{5}}$. Using the binomial expression we can easily do it.

$$\mathrm{(x+y)^{5}=5_{\sum_{r=0}}x^{5-r}.y^{r}}$$

$$\mathrm{(x+y)^{5}=^5{C_{0}}x^{5}.y^{0}+^5{C_{1}}x^{4}.y^{1}+^5{C_{2}}x^{3}.y^{2}+^5{C_{3}}x^{2}.y^{3}+^5{C_{4}}x^{1}.y^{4}+^5{C_{5}}x^{0}.y^{5}}$$

After calculating all the values of $\mathrm{^n{C_{r}}}$ and putting their values in the expression we get,

$$\mathrm{(x+y)^{5}=x^{5}+5x^{4}y+10X^{3}y^{2}+10X^{2}y^{3}+5x^{1}y^{4}+y^{5}}$$

In this way, using binomial theorem we can expand any expression to the finite power in the form of an equation using the binomial formula.

We are going to discuss in detail about the corollaries of binomial theorems and their implementation in C++ to solve various problems related to algebra and combinatorics.

Corollaries of Binomial theorem

Corollary 1

This corollary of binomial theorem says that sum of all the coefficients in the binomial expansion of any expression is equal to $\mathrm{ 2^{n}}$, where n is the power of the expression.Mathematically, it can be written as:

$\mathrm{n_{\sum_{r=0}}^n{C_{r}}=2^{n}}$,for any positive value of n

Proof

We all know that general equation of binomial expansion is,

$$\mathrm{(a+b)^{n}=^n{C_{r=0}}a^{n-0}b^{0}+^n{C_{r=1}}a^{n-1}b^{1}+........+^n{C_{r=n-1}}a^{1}b^{n-1}+^n{C_{r=n}}a^{0}b^{n}}$$

$$\mathrm{(1+1)^{n}=^n{C_{r=0}}1^{n-0}1^{0}+^n{C_{r=1}}1^{n-1}1^{1}+........+^n{C_{r=n-1}}1^{1}1^{n-1}+^n{C_{r=n}}1^{0}1^{n}}$$

$$\mathrm{=^n{C_{r=0}}+^n{C_{r=1}}+........+^n{C_{r=n-1}}+^n{C_{r=n}}}$$

$$\mathrm{=^n{\sum_{r=0}}^n{C_{r}}=2^{n}.}$$

Using this corollary, we can get the sum of all the coefficients of any expression raised to finite power.

Now let’s see how we can implement this corollary of the binomial theorem in C++ to find the sum of all the coefficients in binomial expansion of any expression.

• To find the sum of all the coefficients of binomial expansion of the expression $\mathrm{(x\:+\:y)^{n}}$, we will take n as input.

• To calculate sum of all the coefficients we just need to find the value of $\mathrm{2^{n}}$. This will be our required answer.

The C++ code to find sum of coefficients using this corollary:

Example

#include <bits/stdc++.h>

using namespace std;

//function to calculate the sum of coefficients
long long int sumofcoefficients(int n){
   long long int a = pow(2,n);
   return a;
}

int main()
{
   int n=18;  //(x+y)^n
   
   cout<<"The sum of all the coefficients in binomial expansion of (x+y)^n "<<sumofcoefficients(n)<<endl;
   

   return 0;
}

Output

The sum of all the coefficients in binomial expansion of (x+y)^n 262144

Time Complexity : O(1)

Space Complexity : O(1)

Corollary 2

This corollary defines the relation between sum of r and rth binomial coefficient from r=0 to r=n. This corollary says that the sum of product of binomial coefficients is equal to $\mathrm{n*2^{n-1}}$.

This corollary can be expressed as:

$$\mathrm{n_{\sum_{r=0}}(r*^n{C_{r}})=n*2^{n-1}}$$

Proof

$$\mathrm{n_{\sum_{r=0}}(r*^n{C_{r}})=n_{\sum_{r=0}}(^r{C_{1}}*^n{C_{r}})=^n{\sum_{r=0}}(\frac{r!}{(r-1)!1!}*\frac{n!}{(n-r)!r!})}$$

This can be further simplified as,

$$\mathrm{n_{\sum_{r=0}}=(\frac{n*(n-1)!}{(r-1)!(n-1-(r-1)!)})=n*n_{\sum_{r=0}}n-1_{C_{r-1}}=n*2^{n-1}.}$$

We can find the sum of product of r and rth binomial coefficients for any positive value of n where r is greater than or equal to 0 and r is less than or equal to n using this corollary.

C++ implementation of the binomial theorem corollary:

• If we want to find the sum of the product of r and rth binomial coefficients where r ranges from [0,n], we can use any positive integer n.

• We will use the above corollary to calculate it.

The C++ code to use the above corollary:

Example

#include<bits/stdc++.h>

using namespace std;

//function to calculate sum of product of r and rth binomial coefficients
long long int sum(int n){
   long long int a=0; 
   
   a = n * pow(2,n-1);  //using corollary of binomial theorem 
   
   return a; // return the sum
}

int main()
{
   int n;
   
   n=9;
   cout<<"The sum of product of r and rth binomial coefficients is "<<sum(n)<<endl;
   

   return 0;
}

Output

The sum of product of r and rth binomial coefficients is 2304

Time Complexity : O(1)

Space Complexity : O(1)

Conclusion

The concepts of binomial theorem and the general binomial expansion of any expression was discussed in the article. We have discussed a few properties of the binomial expressions and the corollaries of binomial theorems and how they are used to solve problems in algebra and their implementation in C++ to ease the process of solving.

I hope after reading this article you have cleared the concept of binomial expansion and corollaries of binomial theorem and their implementation.