1.Algebra Booster

Binomial Theorem 6.33
Co-efficient of x n in the expansion of
96. We have,
1
1+
4x
= (1 + 4x) –1 2
Co-efficient of x n in the expansion of
97. We have,
1
2
1- 9x
+ 20x
1
1-
3x
= 1- 4 x + (- 4 x) + + (- 4 x) n +
1
1+
4x
1
=
1 - (4 + 5) x + 20x
1
=
(1 - 4 x)(1 -5 x)
5 4
= -
(1 -5 x) (1 - 4 x)
Co-efficient of x n in the expansion of
1
2
= 5.5 n – 4.4 n
1- 9x
+ 20x
= 5 n–1 – 4 n+1
98. We have,
1
2
1 -( a + b)
x + abx
1
=
(1 - ax)(1 -bx)
1 Ê a b ˆ
= -
a -b Á
Ë1- ax 1-bx˜
¯
1 [ (1 )
-1 (1 )
-1
= a - ax -b -bx
]
a - b
1
\ Co-efficient of x n (
n n
= a◊a
-b◊b
)
a-
b
n+ 1 n+
1
a - b
=
a-
b
99. We have,
Ê1
+ xˆ
Á
Ë1
- x˜
¯
2
= 3 n .
= (–1) n 4 n .
= (1 + x)(1 – x) –1 2 3 1
(1 )(1 n - n
= + x + x + x + x + + x + x + )
\ Co-efficient of x n = 1 + 1 = 2
100. We have,
2
Ê1
+ xˆ
Á
Ë1
- x˜
¯
= (1 + x) 2 (1 – x) –2
= (1 + 2x + x 2 )[1 + 2x + 3x 2 + 4x 3 + …
+ (n – 1)x n–2 + nx n–1 + (n + 1)x n + …]
\ Co-efficient of x n = n – 1 + 2n + n + 1
= 4n
101. As we know that the general term in the expansion of
(1 – x) –n is
nn ( + 1)( n+ 2)…( n+ r-1)
r
tr
+ 1 = ¥ x
r!
Thus, the co-efficient of x n
1Ê1 ˆÊ1 ˆ Ê1
ˆ
Á + 1 + 2 … + n -1
2Ë ˜Á
2 ¯Ë
˜
2 ¯
Á
Ë
˜
2 ¯
= ¥ (2)
n!
1.3.5…(2 n - 1) n
= ¥ (2)
n
2 ¥ n!
1.3.5…(2 n - 1)
=
n!
102. Let t r+1
th term be the negative term.
We have
nn ( 1)( n 2)…( n ( r 1)) 3
t r+1
= - - - - ¥Á Ê x
ˆ
r! Ë
˜
4 ¯
Thus, t r+1
is negative when
(n – r + 1) < 0
fi 13 - r + 1<
0
3
fi
16 1
r > = 5
3 3
fi r = 6
103. We have,
y = x – x 2 + x 3 – x 4 + …
fi –y = x – x 2 + x 3 – x 4 + …
fi 1 – y = 1 – x + x 2 – x 3 + x 4 – …
fi 1 – y = (1 + x) –1
fi
1
1 - y =
(1 + x)
fi
1 x
y = 1 - = 1 + x 1 + x
104 We have,
y = 2x + 3x 2 + 4x 3 + …
fi 1 + y = 1 + 2x + 3x 2 + 4x 3 + …
-2
1
fi 1 + y = (1- x)
=
2
(1 - x)
fi
fi
2 1
(1 - x)
=
1 + y
1 - x =
1
1 + y
1
fi x = 1 -
1 + y
105. We have,
(1 + x) n nn ( -1) 2 nn ( -1)( n-
2) 3
= 1 + nx + x + x +
2! 3!
n
r