Any algebraic expression consisting of only two terms is
known as a binomial expression. It's expansion in power of x is known as the
e.g. (i) a + x (ii) a2 + (iii) 4x – 6y
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 .
x+nC2 an-2 x2 + . . . + nCr
an–r xr + . . . + nCnxn.
where nC0 , nC1
, nC2 , . . . , nCn are called
Binomial co-efficients. The value of nCr is defined as
Similarly (a – x)n = nC0an–nC1an–1
x+nC2 an-2 x2 – . . . +(–1)r
nCr an–r xr+ . . . +(–1)n
= x7 + 7x5
+ 21x3 + 35 x + + ++ .
General Term in the
The general term in
the expansion of ( a+x)n is (r+1)th term given as tr+1
= nCr an-r xr Similarly the
general term in the expansion of ( x + a)n given as tr+1
= nCr xn-r ar. The terms are
considered from the beginning. .
The (r + 1)th
term from the end = ( n – r + 1)th term from the beginning . .
Coefficient of xr in expansion
of (1 + x)n is nCr .
co-efficient of x24 in .
Since, general term
((r+1) th term) in
x30–2r = 15Cr
If this term
contains x24. Then .
30–3r = 24
&⇒ 3r = 6
&⇒ r = 2
co-efficient of x24 = 15C2 ´9a2.
If the co-efficient
of (2r+4)th term and (r–2) th term in the expansion of (1+x)18 are
equal, find r .
of (2r+4) th term in (1+x)18 = 18C2r+3
(r–2) th term = 18Cr–3
&⇒ 18C2r+3 = 18Cr–3
&⇒ 2r+3+r–3 = 18
&⇒ 3r = 18
&⇒ r = 6
Coefficients of Equidistant
Terms from Begnning and end:
coefficients in the expansion of (a+x)n equidistant from the
beginning and the end are equal. .
n is even
Middle term of the expansion is term.
n is odd
In this case term and term are the middle
e.g. Middle term in the expansion of (1
+ x)4 and (1 + x)5.
Expansion of (1
+ x)4 has 5 terms, so third term is middle term i.e. term.
Expansion of (1 + x)5 has 6
terms, so 3rd and 4th both are middle terms, that is, th and th terms are middle
Find the middle
term in the expression of ( 1- 2x + x2)n.
( 1- 2x + x2)n
= [ ( 1- x)2]n = ( 1- x)2n
Here 2n is even
integer, therefore, th i.e. (
n+1) th term will be the middle term
Now (n+1) th term
in (1 - x)2n = 2nCn (1)2n-n(-x)n
Prove that middle
term in the expansion of is .
Since 2n is even,
therefore th i.e. (n+1)th term
will be the middle term.
Now (n+1) th term
i.e. middle term in is given
tn+1 = 2nCnx2n-n=2nCn
To determine the greatest coefficient in the
binomial expansion of (1 + x)n , consider the following
the (r + 1)th binomial coefficient will be greater than the rth
binomial coefficient when,
But r must be an integer, and therefore when
n is even, the greatest binomial coefficient is given by the greatest value of
r , consistent with (1) i.e., r = and hence the greatest
binomial coefficient is nCn/2.
Similarly if n be odd, the greatest binomial
coefficient is given when,
Show that the
greatest coefficient in the expansion of is .
Since middle term
has the greatest coefficient,
coefficient = coefficient of middle term
To determine the
numerically greatest term (absolute value) in the expansion of (a + x)n,
when n is a positive integer. Consider.
Thus |Tr + 1|
> |Tr| if
. . (2)
will be the greatest term if, r has the greatest value consistent with
Find the greatest
term in the expansion of (2 + 3x)9 if x = 3/2.
≥ Tr if,
90 - 9r ≥ 4r
&⇒ 90 ≥
, r being an integer, hence r = 6.
Tr+1 = T7
= T6+1 = 9C6 (2)3 (3x)6
Find the greatest
term in the expansion of .
Let rth term be
the greatest term.
Since , now
. . . (1)
Now again£ 1
&⇒ . . . (2)
From (1) and (2) follows that
&⇒ r = 8 is the greatest term and
= = .
Properties of Binomial
There are (n + 1)
terms in the expansion of (a+b)n, the first and the last term being
an and bn respectively.
If in the expansion of ( 1+x)43,
the coefficient of (2r+1)th term is equal to the coefficient of (r+2)th term
Given in the
expansion of (1+ x)43, the coefficient of (2r +1)th term = the
coefficient of (r +1)th term so, 43C2r = 43Cr+1
&⇒ 2r + r + 1 = 43 or r = 14.
for the binomial expansion (a + x)n.
triangle is way of directly calculating binomial coefficients for different
indices. It is given as follows.
How to construct a
Pascal's triangle is explained below
The coefficients in
the Binomial Expansion can be easily determined with the help of Pascal's
(ax + b)n = 0 if m > n where Dm is mth
derivative w. r. t. x . .
- Dm (ax + b)n = ann!
if n = m
Dm (ax + b)n