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 ax^{2}
+ 2bx + c ≥ 0 for each x ∈ R if b^{2} –ac £ 0
and a > 0.
Since 2^{n} –1 >
0, The inequality in (1) will hold if and only if
Example.4
If (1 +x)^{n}
= C_{0} + C_{1}x + C_{2}x^{2} +.. . . . +C_{n}x^{n}.
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}
= C_{0}+C_{1}x+C_{2}x^{2} + …+C_{n}x^{n} …(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 x^{n1}
in (1+x)^{n} [n(1+x)^{n1}]
= coefficient of x^{n1}
in n (1+x)^{2n1} = n. (^{2n1}C_{n1}).
Thus, from (2) we
get
=
&⇒
&⇒
(b) We have
=
Since
=
&⇒ 2
= n
= = 2
&⇒ =
\
= + 2
Example.5
Show that ^{n}C_{0} –^{n}C_{1}+^{n}C_{2}+ . . . .
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
= =