Example.1
Prove that 
+ 
+
+ -------- + 
= 
.
Solution
LHS = 
+
+ 
+
-------- + 
+
+ 
2
+ -------- + n
= n + (n–1) + (n–2)
+ …. …..(1).
= 
= RHS
Example.2
Find the sum of the series
Solution
The given series
Now, 
=
Similarly, 
etc.
Hence, the gives
series = 
= 
= 
Example.3
If 
,
prove that for n ≥ 2, 
.
Solution
For x ∈ R, we have 
….
(1)
We know that ax2
+ 2bx + c ≥ 0 for each x ∈ R if b2 –ac £ 0
and a > 0.
Since 2n –1 >
0, The inequality in (1) will hold if and only if
Example.4
If (1 +x)n
= C0 + C1x + C2x2 +.. . . . +Cnxn.
Find .
(a) 
(b) 
Solution
(a) We have 
&⇒ 
Differentiating both
the sides we get
Putting x = 1 in the
above expression, we get
…(1)
But 
= 
…(2)
Let us now evaluate 
We have (1+x)n
= C0+C1x+C2x2 + …+Cnxn …(3)
Differentiating both
sides w.r.t.x we get.
&⇒ 
…(4)
Replacing x by 1/x
in (3) we get
…(5)
Note that 
= coefficient of constant term
in
[using (4) and (5)]
= coefficient of
constant term in 
= coefficient of xn-1
in (1+x)n [n(1+x)n-1]
= coefficient of xn-1
in n (1+x)2n-1 = n. (2n-1Cn-1).
Thus, from (2) we
get 
= 
&⇒ 
&⇒ 
(b) We have 
= 
Since 
= 
&⇒ 2
= n
= 
= 2
&⇒ =
\
= + 2
Example.5
Show that nC0 –nC1
+nC2
+ . . . 
.
Solution
We
have 
= 
+ 
. . . (1)
We know that 
Differentiating both sides w.r.t y, we get.
- 
Thus 
= 
. . . .(2)
and - 
= - 
.
. . . (3)
Putting these values in (1), we obtain
= 
= 
