The document explains the binomial theorem and binomial expressions, which are algebraic expressions containing two terms. It outlines how to expand a binomial expression for positive integer powers and provides details on the coefficients, middle terms, and specific properties relevant to binomial expansions. Additionally, it includes an example to illustrate the concept of coefficients in arithmetic progression and conditions for identifying middle terms based on whether the exponent is odd or even.

1. Binomial Theorem

2. Binomial Expression
• An algebraic expression containing two terms is called a binomial expression.
• For example, (a + b), (2x – 3y), 𝑥 +
1
𝑦
, 𝑥 +
3
𝑥
,
2
𝑥
−
1
𝑥2 etc. are binomial expressions.

3. BINOMIAL THEOREM FOR POSITIVE INDEX
• Such formula by which any power of a binomial expression can be expanded in the form of a series
is known as Binomial Theorem. For a positive integer n , the expansion is given by
• (a+x)n = nC0an + nC1an–1 x + nC2 an-2 x2 + . . . + nCr an–r xr + . . . + nCnxn = 𝑟=0
𝑛
𝑛𝐶𝑟𝑎𝑛−𝑟𝑥𝑟.
• where nC0 , nC1 , nC2 , . . . , nCn are called Binomial co-efficients. Similarly
• (a – x)n = nC0an – nC1an–1 x + nC2 an-2 x2 – . . . + (–1)r nCr an–r xr + . . . +(–1)n nCnxn
• i.e. (a – x)n = 𝑟=0
𝑛
−1 𝑟𝑛
𝐶𝑟𝑎𝑛−𝑟
𝑥𝑟
• Replacing a = 1, we get
• (1 + x)n = nC0 +nC1x+nC2x2 + . . . + nCr xr + . . . + nCnxn
• and (1 – x)n = nC0 –nC1x+nC2x2 – . . . + (–1)r nCr xr + . . . +(–1)n nCnxn
C
y 
then

4. • Observations:
 There are (n+1) terms in the expansion of (a +x)n.
 Sum of powers of x and a in each term in the expansion of (a +x)n is constant and equal to n.
 The general term in the expansion of ( a+x)n is (r+1)th term given as Tr+1 = nCr an-r xr
 The pth term from the end = ( n –p + 2)th term from the beginning .
 Coefficient of xr in expansion of (a + x)n is nCr an - r xr.
 nCx = nCy  x = y or x + y = n.
 In the expansion of (a + x)n and (a –x)n, xr occurs in (r + 1)th term.

5. • Illustration 3: If the coefficients of the second, third and fourth terms in the
expansion of (1 + x)n are in A.P., show that n = 7.
• Solution: According to the question nC1  nC2  nC3 are in A.P.
•
2𝑛(𝑛−1)
2
= 𝑛 +
𝑛(𝑛−1)(𝑛−2)
6
• n2 – 9n + 14 = 0  (n – 2)(n – 7) = 0  n = 2 or 7
• Since the symbol nC3 demands that n should be  3
• n cannot be 2,  n = 7 only.

7. MIDDLE TERM
• There are two cases
•
(a) When n is even
• Clearly in this case we have only one middle term namely Tn/2 + 1. Thus middle term in the expansion
of (a + x)n will be nCn/2 an/2xn/2 term.
•
• (b) When n is odd
• Clearly in this case we have two middle terms namely 𝑇𝑛+1
2
𝑎𝑛𝑑𝑇𝑛+3
2
. That means the middle terms
in the expansion of (a +x)n are 𝑛𝐶𝑛−1
2
. 𝑎
𝑛+1
2 . 𝑥
𝑛−1
2 and 𝑛𝐶𝑛+1
2
. 𝑎
𝑛−1
2 . 𝑥
𝑛+1
2 .

8. • Illustration 7: Find the middle term in the expansion of 𝟑𝒙 −
𝒙𝟑
𝟔
𝟗
.
• Solution: There will be two middle terms as n = 9 is an odd number. The middle
terms will be
9+1
2
𝑡ℎ
and
9+3
2
𝑡ℎ
terms.
• t5 = 9C4(3x)5
−
𝑥
3
6
4
=
189
8
𝑥
17
• t6 = 9C5(3x)4
−
𝑥
3
6
5
= −
21
16
𝑥
19
.

9. GREATEST BINOMIAL COEFFICIENT
• In the binomial expansion of (1 + x)n , when n is even, the greatest binomial coefficient is given by
nCn/2.
• Similarly if n be odd, the greatest binomial coefficient will be

10. NUMERICALY GREATEST TERM