Properties of Binomial Coefficient
For the sake of
convenience the coefficients nC0 , nC1
, . . . , nCr , . . . ,nCn are
usually denoted by C0 , C1 , . . . , Cr , . .
. ,Cn respectively .
- C0 + C1
+ C2 +. . . . . + Cn = 2n.
- C0 - C1
+ C2 -. . . . . Cn = 0.
- C0 + C2
+ C4 +. . . . . = C1 + C3 + C5 +.
. . . . = 2n-1.
- nCr + nCr-1
= n+1Cr
- r nCr
=n n-1Cr-1
.- If nCx
= nCy, then either x = y or x + y = n.
So, nCr
= nCn - r = 
Some Important results
(i) Differentiating (1 + x)n =
C0 + C1x + C2x2 + . . . + Cnxn
, on both sides we .
have, n(1 + x)n-1
= C1 + 2C2x + 3C3x2 + . . . + nCnxn-1 .
. . (1).
x = 1
&⇒ n2n-1 = C1 +
2C2 + 3C3 + . . . +nCn.
x = -1
&⇒ 0 = C1 - 2C2 +
. . . +(-1)n-1 nCn.
Differentiating
(1) again and again we will have different results.
(ii) Integrating (1 + x)n ,
we have,
+
(where C is a
constant )
Put x = 0, we get C
= -
Therefore 
…(1)
Put x = 1 in (1) we
get
Put x = -1 in (1) we
get, 
(i) Problems
Related to series of Binomial coefficients in which each term is a product of
an integer and a binomial coefficient i.e. in the form k nCr
.
Example.1
If (1+x)n = 
then prove that C1
+ 2C2 + 3C3+ . . . + nCn= n2n-1
Solution
Method (i): By summation
rth term
of the given series, tr = r nCr
&⇒ tr = n n-1Cr-1
Sum of the series =
= 
= 
= n 2n-1 .
Method (ii) By calculus
We have ( 1+ x )n = C0
+ C1 + C2 x2 + . . . + Cnxn
…(1).
Differentiating
(1) with respect to x
n(1 +x )n-1
= C1 +2C2x + 3C3 x2 + . . . + n
Cnxn-1 …(2).
Putting x = 1 in
(2)
n 2n-1 =
C1 + 2C2 + . . . + nCn .
(ii) Problems
related to series of binomial coefficients in which each term is a
binomial coefficient divided by an integer i.e. in the form of 
.
Example.2
If (1+ x)n
= 
, show that C0
+ 
Solution
Method (i): rth term of the given series
tr =
Required sum = 
= 
= 
= 
.
Method (ii): By calculus
(1+x)n =
C0 + C1x + C2x2 + . .
+ Cnxn . . . . (1).
Integrating both
the sides of (1) with respect to x between the limits 0 to x
&⇒ 
. . . . (2)
Substituting x = 1
in (2) , we get
.
(iii) Problem
related to series of binomial coefficients in which each term is a product of
two binomial coefficients.
Example.3
If ( 1+x)n = 
then prove that
C0C1
+ C1C2 + . . . . + Cn-1 Cn = 
.
Solution
We have
(1+x )n
= C0 +C1 x +C2x2 + . . + Cnxn
…. (1).
and ( x+1)n
= C0xn +C1xn-1 + . . . + Cn ….(2).
multiplying (1)
and (2), we get
(1+x )2n
= (C0 +C1 x +C2x2 + . . + Cnxn
) (C0xn +C1xn-1 + . . . + Cn).
Equating the coefficient
of xn+1, we get
C0C1
+C1C2 + . . . + Cn-1Cn = 2nCn+1
= 
.
Example.4
If ( 1+x)n
= 
then prove that 
.
Solution
We have
(1 – x)2n = 2nC0
- 2nC1 x + 2nC2 x2 - . .
. . + (-1 )2n. 2nC2nx2n ….(1).
and also ( x + 1)2n = 2nC0
x2n + 2nC1 x2n-1 +. . . + 2nC2n
….(2).
multiplying (1) and (2), we get (2nC0
– 2nC1 x + 2nC2 x2 + . .
. . + ( -1 )2n. 2nC2nx2n )( 2nC0
x2n + 2nC1 x2n-1 +. . . + 2nC2n)
= ( 1 – x2)2n equating the coefficient of x2n,
we get .
(iv) Questions involving alternate binomial
coefficients:
Example.5
Evaluate C0 - C1 + C2 -
C3 +...+ (-1)nCn.
Solution
Here
alternately +ve and - ve sign occur
This
can be obtained by putting (-1) instead of 1 in place of x in
(1
+ x)n = C0 + C1x +...+ nCnxn,
we get C0 - C1 +...+ (-1)nCn = 0.
Now
to obtain the sum C0 + C2 + C4 + ...we add (1
+ 1)n and (1 - 1)n.
Similarly,
the cube roots of unity may be used to evaluate
C0
+ C3 + C6 + ... OR C1 + C4 +... OR
C2 + C5 +...
put
x = 1, x = w, x = w2 in
(1
+ x)n = C0 + C1x +...+ Cnxn
and add to get C0 + C3 + C6 +... the other two
may be obtained by suitably multiplying (1 + w)n and .
(1
+ w2)n by w and w2 respectively.
Example.6
Prove that C0 + C4 + C8
+ L = 
Solution
Since (1+i)n = C0 + C1i
+ C2i2 + C3i3 + L +
Cnin
= (C0 - C2
+ C4 -L ) + i(C1 - C3
+ L)
But (1 + i)n =
Hence 
= C0
- C2
+ C4 + L - - - (1)
Also 2n-1 = C0 + C2
+ C4 + L - - - (2)
Adding (1) and (2) we have finally
C0 + C4 + C8 + L = 
(v) Questions involving the greatest integer function:
These
questions generally involve working with binomial expansions on surds.
Example.7
If I is integral part of (2 + 
)n and f is
fraction part of (2 + )n,
then prove that (I + f) (1 –f) = 1. Also prove that I is an odd Integer.
Solution
(2 + Ö3)n = I + f where I is an
integer and 0 £ f < 1 show that I is odd and that (I + f) (1 - f)=1
Here note that (2 - Ö3)n (2 +Ö3)n
= (4 - 3)n = 1
\ (2 + Ö3)n (2 -Ö3)n
= 1 it is thus required to prove that
(2 - Ö3)n = 1 – f but, (2 - Ö3)n
+ (2 +Ö3)n
= [2n - C1.2n - 1.Ö3 +
C22n - 2..(Ö3)2 - ...]
+ [2n + C1.2n - 1.Ö3 +
C22n -2..(Ö3)2 - ...] = 2[2n
+ C2.2n - 2.3+C42n - 4.32
+ ...].
= even integer
\ Now 0 < (2 - Ö3) < 1
\ 0 < (2 - Ö3)n < 1
\ if (2 - Ö3)n = f '
then I + f + f ' = Even
Now O £f < 1 and 0 < f ' < 1 .
. . (1).
Also I + f + f ' = Even integer
\ f + f ' = Even integer . . .
(2).
(1) and (2) imply that f + f ' = 1 (Q 0
< f + f ' < 2)
\ I is odd and f ' = 1 - f
&⇒ (I + f) (1 - f) =
1.
