BINOMIAL THEORIEM

ARJUNA JEE AIR D1 (2026)
BINOMIAL THEOREM
BINOMIAL THEOREM

1. BINOMIAL EXPRESSION:
Any algebraic expression which contains two dissimilar terms is called binomial expression.
1 1 1
For example : x – y, xy + , –1, +3 etc.
x z (x − y)1/3
2. BINOMIAL THEOREM:
The formula by which any positive integral power of a binomial expression can be expanded in the form
of a series is known as BINOMIAL THEOREM.
If x, y R and n  N, then:
n
(x + y)n = nC0xn + nC1xn–1 y + nC2 xn–2 y2 + ..... + nCrxn–ryr + ..... + nCnyn =  n Cr x n −r yr
r =0

This theorem can be proved by induction.
Observations:
(a) The number of terms in the expansion is (n + 1) i.e. one more than the index.
(b) The sum of the indices of x & y in each term is n.
(c) The binomial coefficients of the terms (nC0, nC1.....) equidistant from the beginning and the end are
equal. i.e. nCr= nCr–1

()
(d) Symbol nCr can also be denoted by n , C(n, r) or A nr .
r
Some important expansions:
(i) (1 + x)n = nC0 + nC1x + nC2x2 + ........ + nCnxn.
(ii) (1 – x)n = nC0 – nC1x + nC2x2 + ........ + (–1)n . nCnxn.
Note: The coefficient of xr in (1 + x)n = nCr& that in (1–x)n = (–1)r .nCr


Illustration 1: Expand : (y + 2)6.
Solution: 6C0y6 + 6C1y5.2 + 6C2y4.22 + 6C3y3.23 + 6C4y2. 24 + 6C5y1 . 25 + 6C6 . 26.
= y6 + 12y5 + 60y4 + 160y3 + 240y2 + 192y + 64.


7
 2y2 
Illustration 2: Write first 4 terms of 1 − 
 5 
2 3
 2y2  7 7  2y2  7  2y2 
Solution: C0, C1  −
7 7
 , C0, C2  −  , C3  − 
 5   5   5 
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, BINOMIAL THEOREM

Illustration 3: If in the expansion of (1 + x)m (1 – x)n, the coefficients of x and x2 are 3 and – 6 respectively
then m is - [JEE 99]
(A) 6 (B) 9 (C) 12 (D) 24
 (m)(m− 1).x 2   n(n − 1) 2 
Solution: (1 + x)m (1 – x)n = 1 + mx + + ...... 1 − nx + 2 x + ......
 2   
Coefficient of x = m – n = 3 ........(i)
n(n + 1) m(m− 1)
Coefficient of x2 = –mn + + = -6 ........(ii)
2 2
Solving (i) and (ii), we get
m = 12 and n = 9.


Do yourself - 1:
5
 x
(i) Expand  3x 2 −  (ii) Expand (y + x)n
 2


Pascal's Triangle:
(x + y)0 1
(x + y)1 x+y
(x + y)2 x2 + 2xy + y2
(x + y)3 x3 + 3x2y + 3xy2 + y3
(x + y)4 x4 + 4x3y + 6x2y2 + 4xy3 + y3
Pascal's triangle
(i) Pascal's triangle - A triangular arrangement of numbers as shown. The numbers give the binomial
coefficients for the expansion of (x + y)n. The first row is for n = 0, the second for n = 1, etc. Each
row has 1 as its first and last number. Other numbers are generated by adding the two numbers
immediately to the left and right in the row above.
(ii) Pascal triangle is formed by binomial coefficient.
(iii) The number of terms in the expansion of (x + y)n is (n + 1) i.e. one more than the index.
(iv) The sum of the indices of x & y in each term is n.
(v) Power of first variable (x) decreases while of second variable (y) increases.
(vi) Binomial coefficients are also called combinatorial coefficients.
(vii) Binomial coefficients of the terms equidistant from the beginning and end are equal.
(viii) rth term from the beginning in the expansion of (x + y)n is same as rth term from end in the
expansion of (y + x)n.
(ix) rth term from the end in (x + y)n is (n – r + 2)th term from the beginning.


3. IMPORTANT TERMS IN THE BINOMIAL EXPANSION:
(a) General term: The general term or the (r +1)th term in the expansion of (x + y)n is given by
Tr+1 = nCrxn–r yr.

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, BINOMIAL THEOREM
11
 1 
Illustration 4: Find: (a) The coefficient of x7 in the expansion of  ax 2 + 
 bx 
11
 1 
(b) The coefficient of x–7 in the expansion of  ax − 2 
 bx 
Also, find the relation between a and b, so that these coefficients are equal.
11
 1 
Solution :(a) In the expansion of  ax 2 +  , the general term is :
 bx 
a11−r
r
 1 
Tr+1 = Cr (ax )   = 11Cr. r .x22–3r
11 2 11–r

 bx  b
putting 22 – 3r = 7
 3r = 15 r = 5
a6 7
 t6 = 11c5 .x
b5
11
 1 
Hence the coefficient of x7 in  ax 2 +  is 11C5a6b–5. Ans.
 bx 
Note that binomial coefficient of sixth term is 11C5.
11
 1 
(b) In the expansion of  ax − 2  , general term is :
 bx 
a11−r
r
 −1 
Tr+1 = 11Cr(ax)11–r  2  = (–1)r 11Cr r .x11–3r
 bx  b
putting 11 – 3r = –7
 3r = 18 r = 6
a 5 –7
 T7 = (–1)6. 11C6 .x
b6
11
 1 
Hence the coefficient of x–7 in  ax − 2  is 11C6a5b–6. Ans.
 bx 
Also given :
11 11
 1   1 
Coefficient of x in  ax 2 +  = coefficient of x–7 in  ax − 2 
7

 bx   bx 
–5 –6
 11C5a b = C6a b
6 11 5


 ab = 1 (11C5 =11C6)
which is the required relation between a and b. Ans.

Illustration 5: Find the number of rational terms in the expansion of (91/4 + 81/6)1000.
Solution: The general term in the expansion of (91/4 + 81/6)1000 is
1000− r r
 1  1  1000 1000− r r
Tr+1= Cr  9 4 
1000
   8 6  = Cr 3 2 2 2
 
   
The above term will be rational if exponents of 3 and 2 are integers
1000 − r r
It means and must be integers
2 2
The possible set of values of r is {0, 2, 4, .........., 1000}
Hence, number of rational terms is 501 Ans.
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, BINOMIAL THEOREM

(b) Middle term:
The middle term(s) in the expansion of (x + y)n is (are) :
(i) If n is even, there is only one middle term which is given by T(n+2)/2= nCn/2. xn/2. yn/2
(ii) If n is odd, there are two middle terms which are T(n+1)/2& T[(n+1)/2]+1
Important Note:
Middle term has greatest binomial coefficient and if there are 2 middle terms their coefficients will be
equal.
n
When r = if n iseven
 nCr will be maximum 2
n −1 n +1
When r = or if n isodd
2 2
 The term containing greatest binomial coefficient will be middle term in the expansion of (1 + x)n
9
 x3 
Illustration 6: Find the middle term in the expansion of  3x − 
 6 
9
 x3 
Solution: The number of terms in the expansion of  3x −  is 10(even). So there are two middle terms.
 6 
th th
 9 +1  9+3
i. e.   and   are two middle terms. They are given by T5 and T6
 2   2 
4
 x3  x12 9.8.7.6 35 17 189 17
 T5 = T4+1 = C4(3x)  −  = 9C435x5. 4 =
9 5
. x = x
 6  6 1.2.3.4 24.34 8
5
 x3  x15 −9.8.7.6 34 19 21
and T6 = T5+1 = C5(3x)  −  = – 9C434.x4. 5 =
9 4
. 5 5 x = – x19 Ans.
 6 6 1.2.3.4 2 .3 16

(c) Term independent of x:
Term independent of x does not contain x ; Hence find the value of r for which the exponent of x is
zero.
10
 x  3 
Illustration 7: The term independent of x in  +  2   is –
 3  2x  
5
(A) 1 (B) (C) 10C1 (D) none of these
12
Solution: General term in the expansion is
r 10 − r
35−r
3r
 x 2  3  2 −10 3r 20
10
Cr    2 = 10Cr x 2 . 10− r
For constant term, = 10  r =
3  2x  2 3
2 2
which is not an integer. Therefore, there will be no constant term. Ans. (D)

Do yourself - 2:
10
 1
(i) Find the 7th term of  3x 2 − 
 3
25
 3
(ii) Find the term independent of x in the expansion:  2x 2 − 3 
 x 
6 7
 2x 3   1
(iii) Find the middle term in the expansion of: (a)  −  (b)  2x 2 − 
 3 2x   x

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