Example.1
If n is an in integer greater than 1, prove
that
a–C1(a–1)+C2(a–2)
– . . . + (–1)n Cn (a–n) = 0.
Solution
LHS = a –C1(a–1) + c2 (a–2) –. . .+ (–1)n
Cn (a–n).
= a [1–C1+C2
– . . . (–1)n Cn] + [C1–2C2+3C3–
. . .+(–1)n–1 nCn].
= aP + Q .
. . .(1).
P = C0 –C1
+ C2 – . . . + (–1)n Cn = 0 . . . (2).
Also,
C0 + C1x
+ C2x2 + . . . + Cnxn = (1+x)n.
Differentiating
w.r.t. x,.
C1 + 2C2x
+ 3C3x2 + . . . . + nCnxn-1 =
n(1+x)n–1.
Putting x = –1
C1 –2C2
+ 3C3 –. . . + (–1)n–1 nCn= 0
. . . (3).
LHS = aP + Q = a´
0 + 0 = 0. .
Example.2
If 
and ak = 1 for all k ≥n. Then show that
bn = 2n+1Cn + 1 .
Solution
Put x - 3 = y, we
get 
. We have to find coefficient of
yn in Left hand side as coefficient of yn in the right
hand side is bn. And we will start getting coefficient of yn
only when r ≥ n in the left hand
side.
Coefficient of yn
in (1 + y)n + (1 + y)n + 1 + ... + (1 + y)2n =
bn.
Coefficient of yn
in (1 + y)n 
&⇒ Coefficient of yn + 1 in 
= bn
&⇒ 2n+1Cn+1 = bn
.
Example.3
If (1 + x)n = 
show that 
Solution
rth factor of 
is
given by
tr = 
Now =
t1 . t2 . t3 …… tn
= 
= 
= 
Example.4
Prove that 
= (–1)n–1 nCn
Solution
(1+x)2n =
2nC0 + 2nC1x+2nC2x2
+ . . . + 2nC2nx2n . . . (1).
Differentiating w.r.t. x, we get .
2n(1+x)2n–1 = C1 +2C2x
+ 3C3x2+ . . . +2nC2nx2n-1 .
. . (2).
Also 
=
C0 – C1 
+C2
. . . + C2n 
. . . (3)
Where Cr = 2nCr
Multiplying (2) and (3), we get,
2n(1+x)2n–1
= (C1+2C2 x +3C3x2
+ . . . +2nC2n x2n-1) (C0 –C1+C2. . . C2n )
The coefficient of 
in RHS is
= 
Also, the coefficient of in 2n(1+x)2n-1
= coefficient of in
2n. 
= coefficient of in
(1–x2)2n–1(1–x)
= coefficient of x2n–2 in 2n (1–x2)2n–1(1–x)
= coefficient of x2n-1 in 2n (1–x2)2n–1
´ coefficient of x
in(1–x)
= coefficient of (x2)n–1
in 2n (1–x2)2n–1 ´
coefficient of x in (1–x)
= 2n (-1)n–1 2n–1Cn–1(-1)
= (-1)n 2n 
= (-1)n 
n = (-1)n 
n = - (- 1)n–1 nCn ….
(6)
From (5) and (6), we get, 
= (-1)n–1 nCn
Example.5
Given sn = 1 + q +q2+........+qn.
Sn = 
Prove that (n+1)C1
+ (n+1)C2s1 + (n+1)C3s2+........+(n+1)Cn+1sn
= 2n Sn.
Solution
sn is in geometric progression,
hence
Consider (n+1)C1
+ (n+1)C2s1 + (n+1)C3s3+........+(n+1)Cn+1sn.
= 
Since 
