1.Algebra Booster

Binomial Theorem 6.35
109. Let p = q + h
p h
fi = 1 +
q q
We have,
5p
+ 4q
4p
+ 5q
Ê pˆ 5Á
+ 4
Ë q
˜
¯
=
Ê p ˆ
4Á
+ 5
Ë q
˜
¯
Ê hˆ
51 Á + + 4
Ë q
˜
¯
=
Ê h ˆ
4Á1+ + 5
Ë q
˜
¯
Ê 5hˆ
Á1
+
Ë 9q
˜
¯
=
Ê 4h
ˆ
Á1
+
Ë 9q
˜
¯
110. Let x = 1 + h
We have,
m
mx
m
Ê 5hˆ Ê 4hˆ
= Á1+ ¥ 1+
Ë 9q
˜
¯
Á
Ë 9q
˜
¯
Ê 5hˆ Ê 4hˆ
= Á1+ ¥ 1-
Ë 9q
˜
¯
Á
Ë 9q
˜
¯
5h
4h
= 1 + -
9q
9q
h Ê hˆ
= 1+ = 1
9q
Á +
Ë q
˜
¯
Ê pˆ
= Á
Ë q ˜
¯
-
-
n
1/9
1/9
-1
nx
n
m
n
m(1 + h) - n(1 + h)
=
m-
n
m(1 + mh) - n(1 + nh)
=
m-
n
2 2
( m- n) - ( m - n ) h
=
m-
n
= 1 – (m + n)h
= (1 – h) (m+n)
= x (m+n)
111. We have,
2 3 4
n
(5 x) (5 x) (5 x) (5 x)
e 5x = 1+ 5x
+ + + + + +
2! 3! 4! n!
Thus,
n
5
the co-efficient of x n in e 5x =
n!
112. We have,
e 5x+4
= e 4 .e 5x 2 3
4
Ê
n
5 x (5) x (5) x (5) x
= e Á1
+ + + + + +
Ë 1 2! 3! n!
Thus,
the co-efficient of x n in e 5x+4 4
Ê
n
5 ˆ
= e ¥ Á
Ë n!
˜
¯
113. We have,
Ê
2
1+ 3x
+ 2x
ˆ
Á
Ë x ˜
e ¯
= (1 + 3x + 2x 2 ) ¥ e –x
2
= (1 + 3x
+ 2 x ) ¥
Ê
2 3
n n
x x (-1)
x
Á1
- x + - + + +
Ë 2! 3! n!
Thus,
the co-efficient of x n
n n-1 n-2
(-1) 3 ◊( -1) 2 ◊( -1)
= + +
n! ( n -1)! ( n - 2)!
114. Given,
x
e
2
n
B0 B1x B2x Bn
x
1 - x = + + + +
Now,
e x ¥ (1 – x) –1
Ê
2 3 4
n
x x x x ˆ
= Á1
+ x + + + + + +
Ë 2! 3! 4! n!
˜
¯
2 3 n
¥ (1 + x + x + x + + x + )
Ê 1 1 1 1 ˆ n
= + + + + + 1 x
Ë
Án! ( n -1)! ( n - 2)! 1! ¯
˜
Ê 1 1 1 1 ˆ
+ Á + + + + + 1 x
Ë( n -1)! ( n - 2)! ( n -3)! 1! ˜
¯
ˆ
˜
¯
ˆ
˜
¯
n-1
Thus, comparing the co-efficients of
x n–1 and x n , we get
1 1 1 1
B = n
1
n! + ( n -1)! + ( n - 2)! + + 1!
+
and
1 1 1 1
B n - 1 = 1
( n -1)! + ( n - 2)! + ( n -3)! + + 1!
+
1
Therefore, Bn
- Bn-1
= .
n!
115 We have,
1 1 1 1 1
1 + + + + = ( )
2! 4! 6! 2 e + e-
1 1 1 1 1
and 1 + + + + = ( )
3! 5! 7! 2 e - e-