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

=
= 