# Binomial Theorem for Positive Integral Indices

We all know about these four important expansions: (l+ m)2, (l-m)2, (l + m)3, (l – m)3

• (l +m)2 = l2 + m2 + 2lm

• (l-m)2 = l2 + m2 – 2lm

• (l + m)3  =  l3 + m3 + 3lm(l + m)

• (l – m)3  =  l3 – m3 – 3lm(l – m)

In the above expansions,

1. We see that the total number of terms in the expansion is one more than the index. For example, in the expansion of (l+ m)2, the number of terms is equal to 3 whereas the index of (l+ m)2 is equal to 2.

2.  Powers of the first quantity ‘l’ generally go on decreasing by 1 whereas the powers of the second quantity which is ‘m’ increase by 1, in the successive terms.

3.  In each term of the above expansion, the sum of the indices of l and m is the same and is equal to the index of l+m.

Using the above expansions, we can easily find out the values of,

(102)2 = (100 + 2)2

=1002 + 2 × 100 × 2 + 22

=10000 + 400 + 4 = 10404

Similarly,

(106)2 = (100 + 6)2

=1002+6 × 100 × 2 + 62

=10000 + 1200 + 36 = 11236

But, finding out the values of (103)6, (119)5 with repeated multiplication is difficult. This is the reason why we use binomial Theorem, it makes repeated multiplication easy.

### What is the statement of Binomial Theorem for Positive Integral Indices –

• The Binomial theorem states that “the total number of terms in an expansion is always one more than the index.”

•  For example, let us take an expansion of (a + b)n, the number of terms for the expansion is n+1 whereas the index of expression  (a + b)n is n, where n is any positive integer.

• By using the Binomial theorem, we can expand (x +y)n, where n is equal to any  rational number. Let’s discuss the binomial theorem for positive integral indices.

Now, let us write the expansion of (x+y)n , [0≤n≤5 , where n is an integer] and let’s find the properties of binomial expansion:

(m+n)0=1

(m+n)1=(m+n)

(m+n)2=m2+2mn+n2

(m+n)3=m3+3m2 n+3mn2+ n3

(m+n)4=m4+4m3 n+6m 2n2+4mn3+n4

(m+n)5=m5+5m4 n+10m3 n2+10m2 n3+5mn4+n5

### What is the Binomial Theorem for Positive Integral Indices?

• Binomial Theorem for index as positive integer n is denoted by :

 (a +b)n = nC0​an+ nC1​an-1 b+nC1 an-1 b2 +…………+ nCn-1 ​abn-1+ nCn​bn

• We know that nCr   =$\frac{n!}{r!(n – r)!}$, where n = a non-negative integer and 0≤r≤n. Also , the value of nCn and nC0 is equal to 1.

### Proof of Binomial Theorem –

Using mathematical induction, let us prove binomial theorem:

P(n): (a + b)n = nC0 an + nC1 an-1 b + nC2 an-2 b2 + ………nCn-1 a bn-1 + nCn bn

For n equal to 1,

P(1): (a+ b)1 = 1C0 a1 + 1C1 b1 = a + b, which is true.

Let us suppose that P(k) is true for any positive integer k.

Then, we get,

P(k): (a+ b)k = kC0 ak + kC1 ak-1 b + kC2 ak-2 b2 + ………kCk-1abk-1 + kCkbk…………… Equation (1)

P(k+1) : (a+b)k+1 = (k+1)C0ak+1 +( k+1)C1akb + (k+1)C2ak-1b2 + ……… + (k+1)Ck+1bk+1

(a+b)k+1

(a+b)k+1 = (a+b) (a+b)k

= (a+b) (kC0ak + kC1ak-1b + kC2ak-2b2 + ………kCk-1abk-1 + kCkbk)

kC0ak+1 + kC1akb + kC2ak-1b2 + ………kCk-1a2bk-1 + kCkabk + kC0akb + kC1ak-1b2 + kC2ak-2b3 + ……… kCk-1abk + kCkbk+1

Grouping the like terms, we get

kC0ak+1 + (kC1 + kC0)akb + (kC2 + kC1 )ak-1b2 + ……………… (kC1 +kCk-1 ) abk + kCkbk+1

kC0 =1 = k+1C0kCr + kCr-1 = k+1Cr, and kCk = k+1 Ck+1 = 1

⇒ (a+b)k+1 = k+1C0ak+1 + (k+1)C1akb + (k+1)C2ak-1b2 + ……………… (k+1)Ck abk + (k+1)Ck+1bk+1

P (k+1) is true whenever P(k) is true.

Therefore, P(n) is true for all positive integral values of n.

### Formula for Pascal’s Triangle –

Pascal’s triangle can be defined as a triangular array of binomial coefficients. The structure of the Pascal triangle was developed by a French mathematician, Blaise Pascal.
In the above diagram we see that a particular pattern is followed.

The diagram below shows how we get the values, it shows the flow of how the numbers are added in a Pascal’s triangle.

We notice that in the triangle 3 = 1 + 2, 6 = 3 + 3 and so on.

In a Pascal triangle the terms in each row (n) generally represent the binomial coefficient for the index = n − 1 , where n = row

For example, Let us take the value of n = 5, then the binomial coefficients are  1 ,5,10, 10, 5 , 1.

We know that nCr   =$\frac{n!}{r!(n – r)!}$, where n = a non – negative integer and 0 ≤ r ≤ n.

Also , the value of nCn and nC0 is equal to 1.

The Pascal’s Triangle can be now re-written,

### Questions to be Solved :

Question 1) Prove the binomial theorem.

Solution) Here’s the proof for binomial theorem,

### Proof of Binomial Theorem –

Using mathematical induction, let us prove binomial theorem:

P(n): (a + b)n = nC0 an + nC1 an-1 b + nC2 an-2 b2 + ………nCn-1 a bn-1 + nCn bn

For n equal to 1,

P(1): (a+ b)1 = 1C0 a1 + 1C1 b1 = a + b, which is true.

Let us suppose that P(k) is true for any positive integer k.

Then, we get,

P(k): (a+ b)k = kC0 ak + kC1 ak-1 b + kC2 ak-2 b2 + ………kCk-1abk-1 + kCkbk…………… Equation (1)

P(k+1) : (a+b)k+1 = (k+1)C0ak+1 +( k+1)C1akb + (k+1)C2ak-1b2 + ……… + (k+1)Ck+1bk+1

(a+b)k+1

(a+b)k+1 = (a+b) (a+b)k

= (a+b) (kC0ak + kC1ak-1b + kC2ak-2b2 + ………kCk-1abk-1 + kCkbk)

kC0ak+1 + kC1akb + kC2ak-1b2 + ………kCk-1a2bk-1 + kCkabk + kC0akb + kC1ak-1b2 + kC2ak-2b3 + ……… kCk-1abk + kCkbk+1

Grouping the like terms, we get

kC0ak+1 + (kC1 + kC0)akb + (kC2 + kC1 )ak-1b2 + ……………… (kC1 +kCk-1 ) abk + kCkbk+1

kC0 =1 = k+1C0kCr + kCr-1 = k+1Cr, and kCk = k+1 Ck+1 = 1

⇒ (a+b)k+1 = k+1C0ak+1 + (k+1)C1akb + (k+1)C2ak-1b2 + ……………… (k+1)Ck abk + (k+1)Ck+1bk+1

P (k+1) is true whenever P(k) is true.

Therefore, P(n) is true for all positive integral values of n.