Example.1
Find the larger of + and .
Solution
We have (101)50 = (100+1)50
= 10050 +
50.10049 + 10048
+ --------
And 9950 =
(100–1)50 = 10050 – 50. 10049 + . 10048 - ….
Subtracting we get
(101)50
- 9950 =
= 10050 +
2 > 10050
Hence, (101)50
> 9950 + 10050
Example.2
If the 3rd, 4th, 5th
and 6th terms in the expansion of (x + a)
n be respectively a, b, c and d, prove that .
Solution
Putting r = 3, 4, 5 in the
above, we get
.
. . (1)
. .
. . (2)
.
. ..(3)
Now dividing (1) by 2 and
(2) by 3 respectively, we get
.
. .(4)
.
. .(5)
Subtracting 1 from both
sides of (4) and (5)
.
. .(6)
.
. .(7)
Dividing (6) and (7), we
get
by (5)
Example.3
Find the last three digits
of 17256.
Solution
We have 172 =
289 =290 –1.
Now, 17256 = (172)128
= (290 –1)128
where m is a positive integer.
= 1000 m + (128) (290)
[(127) (145) –1 ] + 1
= 1000 m + (128) (290)
(18414) + 1
= 1000 [m + 683527] + 680 +
1 = 1000 [m + 683527] + 681
Thus, the last three digits
of 17256 are 681.
Example.4
Show that + =
198. Hence show that the integral part of is
197.
Solution
Since (x+a)n + (x–a)n
= 2 (xn + nC2xn–2a2+nC4xn–4a4+nC6xn–6a6
+ ----)
Here, n =6, 6C2 = 15, 6C4
= 15, 6C6 =1 x = ,
a = 1
\ +
= 2
= 2 [8+15.4+15.2+1].
= 2 (99) = 198
\ =
198 – . . . .
(1)
Now, = 1.4–1 = .4 <1
\ <
1 and it is certainty positive
\ 0< < 1 .
. . (2)
Hence, from (1)
= 198 – (some thing
positive but less than1)
Therefore the
integral part of is 197.
Example.5
Find the term
independent of x in the expansion of .
Solution
(1
+ x + 2x3)
=
+ 9C2 - 9C3
- 9C5-
9C7
\
Term independent of x
= 9C6
= =
=