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Binomial Expansion Formula

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Algebra and series are important topics of mathematics. Binomial expansion is an expansion of the binomial expression (two terms algebraic expression). It requires prior knowledge of combinations, mathematical induction. This expansion gives the formula for the powers of the binomial expression. Binomial expansion formula finds the expansion of powers of binomial expression very easily.

Binomial Expansion 

In algebraic expression containing two terms is called binomial expression. Example: (x + y), (2x – 3y), (x + (3/x)). The general form of the binomial expression is (x + a) and the expansion of (x + a)n, n ∈ N is called the binomial expansion. Binomial expansion provides the expansion for the powers of binomial expression.

Binomial Expansion Formula

If x and a are real numbers, for all n ∈ N,

(x + a)n = nC0xna0 + nC1xn-1a1 + nC2xn-2a2 + ………+ nCr xn-rar + …….. + nCn-1x1an-1 + nCnx0an

(x + a)n = nCr xn-rar 

Proof: 

Proof of binomial expansion using the principle of mathematical induction on n.

Let X(n) be : (x + a)n = nC0xna0 + nC1xn-1a1 + nC2xn-2a2 + ………+ nCr xn-rar + …….. + nCn-1x1an-1 +nCnx0an

Step I:- 

To prove: X(1) : (x + a)1 =1C0x1a0 + 1C1x0a1

We know that : (x + a)1 = x + a = 1C0x1a0 + 1C1x0a1

therefore, X(1) is true.

Step II:- let X(m) be true. Then,

(x + a)m = mC0xma0 + mC1xm-1a1 + mC2xm-2a2 + ………+ mCm-1x1am-1 +mCmx0am    ————(1)

To prove :- X(m+1) is true. i.e. 

(x + a)m+1 = m+1C0xm+1a0 + m+1C1xma1 + m+1C2xm-1a2 + ………+ m+1Cmx1am +m+1Cm+1x0am+1

Proof:-  (x + a)m+1 = (x + a)(x + a)m

= (x + a)[mC0xma0 + mC1xm-1a1 + mC2xm-2a2 + ………+mCrxm-rar+ mCm-1x1am-1 +mCmx0am]

= mC0xm+1a0 + (mC1 + mC0)xma1 + (mC2 + mC1)xm-1a2 + … +(mCr + mCr-1)xm-r+1ar + … + (mCm-1 + mCm)x1am + mCmam+1

[Since, mCr-1 + mCr = m+1Cr , r = 1,2,3…..,m]

= m+1C0xm+1a0 + m+1C1xma1 + m+1C2xm-1a2 + ………+ m+1Cmx1am + m+1Cm+1x0am+1

X(m + 1) is true.

X(m) is true ⇒ X(m + 1) is true.  

Important points about the binomial expansion formula

  1. As in Binomial expansion, r can have values from 0 to n, the total number of terms in the expansion is (n+1).
  2. The sum of indices of x and a in each term is n.
  3. Since, nCr = nCn-r , for r = 0,1,2……,n . Hence, the coefficients of terms equidistant from the starting and end are equal. So such coefficients are known as binomial coefficients.
  4. (x-a)n =\sum_{r=0}^{n}     (-1)r nCrxn-ra  In the expansion of (x-a)n we have alternate positive and negative terms and sign of last term depends on the value of n (odd or even).
  5. The coefficient of (r+1)th term or xr in the expansion of (1 + x)n is nCr.
  6. (x + a)n + (x – a)n = 2[nC0xna0 + nC2xn-2a2 + …….] and (x + a)n – (x – a)n = 2[nC1xn-1a1 + nC3xn-3a3 + …….] 
  7. In the binomial expansion of (x + a)n, the general term is given by Tr+1 = nCrxn-rar
  8. In the binomial expansion of (x – a)n, the general term is given by Tr+1 = (-1)r nCrxn-rar
  9. The binomial expansion of (x + a)n contains (n + 1) terms. Therefore, if n is even, then ((n/2) + 1)th term is the middle term and if n is odd, then ((n + 1)/2)th and ((n + 3)/2)th terms are the two middle terms.  
  10. Different values of n have a different number of terms:
  n(x + a)n + (x – a)n(x + a)n – (x – a)n
odd       (n+1)/2       (n+1)/2
even       (n/2)+1        (n/2)

Sample Questions

Question 1: Find the number of terms in the expansions of the following :

(i) (9x – y)    (ii) (1 +7x)9 + (1 – 7x)9      (iii) (1 + 2x + x2)20

Solution: 

(i) In the expansion of (x + a)n the number of terms is (n+1). Hence, in the expansion of (9x – y)9 the number of terms is 10.

(ii) In the expansion of (x + a)n + (x – a)n the number of terms is (n+1)/2 if n is odd. So number of terms in the expansion of (1 + 7x)9 + (1 – 7x)9 = (10/2) = 5   

(iii) (1 + 2x + x2)20 = [(1 + x)2]20 = (1 + x)40. Hence, number of terms = 41

Question 2: Expand (3x + 8)4

Solution: 

According to binomial expansion :

(3x + 8)4 = 4C0 (3x)4 (8)0 +  4C1 (3x)3 (8)1 + 4C2 (3x)2 (8)2 +  4C3 (3x)1 (8)3 4C4 (3x)0 (8)4

= (3x)4 + 4.(3x)3.8 + 6.(3x)2.64 +4.(3x).512 + 4096

= 12x4 + 864 x3 + 3456 x2 + 6144 x + 4096

Question 3: Expand (2x – 1)5

Solution: 

According to binomial expansion :

(2x – 1)5 = (2x + (-1))5 = 5C0 (2x)5 (-1)0 5C1 (2x)4 (-1)1 + 5C2 (2x)3 (-1)2 +  5C3 (2x)2 (-1)3 +  5C4 (2x)1 (-1)4 + 5C5 (2x)0(-1)5

= 32x5 – 5.16x4 + 10.8x3 – 10.4x2 + 10x – 1

= 32x5 – 80x4 + 80x3 – 40x2 + 10x – 1

Question 4: Expand (1 + x + x2)3

Solution: 

Let, y = x + x2

(1 + x + x2)3 = (1 + y)3 = 3C0 (1)3 (y)0 +  3C1 (1)2 (y)1 + 3C2 (1)1 (y)2 +  3C3 (1)0 (y)3 

= 1 + 3y + 3y2 + y3

= 1 + 3(x + x2) + 3(x + x2)2 + (x + x2)3 = 1 + 3x + 3x2 + 3(x2 + x4 + 2x3) + (x3 + x6 + 3x4 + 3x5)

= 1 + 3x +  3x2 + 3x2 + 3x4 + 6x3 + x3 + x6 + 3x4 + 3x

= 1 + 3x + 6x2 + 7x3 + 6x4 + 3x5 + x6

Question 5: Find (a + b)4 – (a – b)4. Hence, evaluate (√3 + √2)4 – (√3 – √2)4.

Solution: 

(a + b)4 – (a – b)4 = 2.[4C1a3b1 + 4C3a1b3] = 2.[4a3b1 + 4a1b3] = 8a3b1 + 8a1b3 

Put a = √3 and b = √2 

(√3 + √2)4 – (√3 – √2)4 = 8.(√3)3(√2) + 8.(√3)(√2)3 = 24√6 + 16√6 = 40√6 

Question 6: Find the 10th term in the binomial expansion of (4x2 + 1/x)11.

Solution: 

In the binomial expansion of (x + a)n , (r+1)th term is given by Tr+1 = nCrxn-rar

In the expansion of (4x2 + 1/x)11 , [n = 11, r = 9, x = 4x2, a = 1/x]

T10 = T9+1 = 11C9 (4x2)11-9 (1/x)9 = 55.(16x4).(1/x9) = 880/x5

Question 7: Find the middle term in the expansion of [(4/3)x2 – (3/4x)]20.

Solution: 

Here, n = 20 (even)

[(20/2) + 1]th term i.e. 11th term is the middle term.

Hence, the middle term = T11 = T10+1 = 20C10.[(4x2/3)]20-10. [-(3/4x)]10 = 20C10.x10



Last Updated : 06 Jan, 2024
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