Example.1
Show that C0 –22C1
+ 32C2 - . . . +(–1)n(n+1)2Cn
= 0, n> 2.
Solution
We have
C0 + C1x
+ C2x2 + . . . + Cnxn = (1+x)n
. . . (1).
Multiplying (1) by
x
C0x + C1x2+C2x3
+. . . +Cnxn+1 = x(1+x)n . . .
(2).
Differentiating (2) w.r.t.
x, we have .
C0 + 2c1x +3C2x2
+ . . . +(n+1) Cnxn = (1+x)n + nx(1+x)n–1
.
= [(n+1)x+1] (1+x)n–1
. . . (3).
Multiplying (3) by
x, we get
C0x + 2C1x2
+ 3C2x3 + . . . + (n+1) Cnxn–1 =
[(n+1)x2+x](1+x)n-1 . . . (4).
Differentiating (4)
w.r.t x , we get .
C0 + 22
C1x + 32 C2x2 + . . . + (n+1)2
Cnxn.
= [2(n+1)x+1)] (1+x)n–1
+ (n–1) [(n+1)x2 + x] ´
(1+x)n–2 . . . (5).
Putting x = –1 in
(5) , we get
C0–22C1
+ 32C2 - . . . +(–1)n (n+1)2 Cn
= 0.
Example.2
Prove that C0Cr + C1Cr+1
+ C2Cr+2 + . . . + Cn–r Cn = 
Solution
We have,
C0 + C1x
+ C2x2 + . . . + Cnxn = (1+x)n
. . . (1).
Also, C0xn
+ C1xn–1+ C2xn–2 + . . . + Cn
= (x+1)n . . . (2).
Multiplying (1) and
(2), we get
(C0+C1x+C2x2
+…+ Cnxn) (C0xn+ C1xn–1
+C2xn–2 + …+Cn)
= (1+x) 2n .
. . (3).
Equating coefficinet
of xn–r from both sides of (3), we get
C0Cr
+ C1Cr+1 + C2Cr+2 + ------------+ Cn–rCn
= 2n Cn–r = 
Example.3
Find the sum of the series
Solution
The given series
Now, 
=
Similarly, 
etc.
Hence, the gives
series = 
= 
= 
Example.4
Let k and n ∈ I+ and Sk = 1k + 2k
+. . . . + nk, show that , m+1C1S1
+ m+2C2S2+ . . . . + m+1CmSm
= (n+1)m+1 – (n+1) . .
Solution
We have
&⇒
Putting p = 1, 2, 3,…,n and adding, we get
&⇒
Example.5
Show that 
.
Solution
We have, for r ≥ 0,
= 
= 
Thus, 
= 
= 
[put r+2 = s]
= 
= 
= 
=
But 2n+3 + (-1)n
= 
Thus, 
