Example.1
Prove that
+
+
+ -------- +
=
.
Solution
LHS =
+
+
+
-------- + ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image009.gif)
+
+
2
+ -------- + n![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image013.gif)
= n + (n–1) + (n–2)
+ …. …..(1).
=
= RHS
Example.2
Find the sum of the series
![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image015.gif)
![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image016.gif)
Solution
The given series
![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image017.gif)
![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image018.gif)
Now,
=![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image020.gif)
Similarly, ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image021.gif)
etc.
Hence, the gives
series = ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image023.gif)
=
= ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image025.gif)
Example.3
If
,
prove that for n ≥ 2,
.
Solution
For x ∈ R, we have ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image028.gif)
….
(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
![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image031.gif)
![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image032.gif)
Example.4
If (1 +x)n
= C0 + C1x + C2x2 +.. . . . +Cnxn.
Find .
(a)
(b) ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image034.gif)
Solution
(a) We have ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image035.gif)
&⇒ ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image036.gif)
Differentiating both
the sides we get
![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image037.gif)
Putting x = 1 in the
above expression, we get
…(1)
But ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image039.gif)
=
…(2)
Let us now evaluate ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image041.gif)
We have (1+x)n
= C0+C1x+C2x2 + …+Cnxn …(3)
Differentiating both
sides w.r.t.x we get.
![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image042.gif)
&⇒
…(4)
Replacing x by 1/x
in (3) we get
…(5)
Note that
= coefficient of constant term
in
![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image046.gif)
[using (4) and (5)]
= coefficient of
constant term in ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image047.gif)
= 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 ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image048.gif)
= ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image049.gif)
&⇒ ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image050.gif)
&⇒ ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image051.gif)
(b) We have ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image052.gif)
= ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image053.gif)
Since ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image054.gif)
=
&⇒ 2![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image054.gif)
= n
=
= 2![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image058.gif)
&⇒
=
![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image059.gif)
\ ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image052.gif)
=
+ 2![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image060.gif)
Example.5
Show that nC0 –nC1
+nC2
+ . . .
.
Solution
We
have ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image064.gif)
= ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image065.gif)
+
. . . (1)
We know that ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image067.gif)
Differentiating both sides w.r.t y, we get.
- ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image068.gif)
Thus ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image069.gif)
=
. . . .(2)
and - ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image071.gif)
= -
.
. . . (3)
Putting these values in (1), we obtain
![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image073.gif)
=
= ![](http://www.quizsolver.com/radix/dth/notif/BIO%20SUB%205_files/image075.gif)