1.Algebra Booster

6.24 Algebra Booster
30 2 30 2 2 30 2 3
24. Since n is odd, so, the middle terms are
Cx 1 C2 x C3
x
30 2 5 30 2 30
Ê9 + + C5( x ) + + C30( x )
+ 1ˆ Ê9 3
th and
+ ˆ
Á ˜ Á ˜th
terms
Ë 2 ¯ Ë 2 ¯
= 5th and 6th terms.
3
4
9 9-
4
Ê x ˆ
Thus, t5= t4+
1= C4
¥ (3 x)
¥ Á-
Ë
˜
6 ¯
9 Ê 3 ˆ 17
= C4
¥ Á ¥ x
Ë
˜
16¯
3
5
9 9-5
Ê x ˆ
and t6 = t5+
1= C5¥ (3 x)
¥ Á-
Ë
˜
6 ¯
16 17
19
9
Ê x ˆ
= C5
¥ Á-
Ë
˜
96 ¯
5 5 2 5 3
= (1 + Cx 1 + C2 x + Cx 3 + )
(2 n)!
n
6 6 2 6 3
= ¥ x
¥ (1 - Cx 1 + C2x - Cx 3 + )
( n)! ¥ ( n)!
n
2 ¥ ( n)! ¥ {1.3.5…(2 n-1)}
= ¥
( n)! ¥ ( n)!
n
2 ¥ {1.3.5…(2 n -1)}
n
= ¥ x
( n)!
m m n n
Cx 1 Cx 2 Cx 1 Cx 2
100/2 100/2
100 Ê1ˆ Ê2ˆ
= C100/2
¥ Á ¥
Ë
˜
2¯ Á
Ë
˜
3¯
50 50
100 Ê1ˆ Ê2ˆ
( -1) nn ( -1)
= C50
¥ Á ¥
Ë
˜
+ - mn =-6
2¯ Á
Ë
˜
3¯
2 2
50
100 Ê1ˆ
= C50
¥Á
Ë
˜
3¯
28. Put a = 1 = b, fi 2 n = 4096 = 2 12
fi n = 12
Thus, the greatest co-efficient is 12 C 6
.
29. Put x = 1, fi 3 n = 6561 = 3 9
fi n = 9
Thus, the greatest co-efficients are
24 ¥ 9 C 4
and 2 5 ¥ 9 C 5
.
30. We have,
Ê ˆ
Á + ˜ =
( n+
1)| x|
Ë 2 ¯
m =
a + | x|
2
12-6 6
12
Ê2x
ˆ Ê 3 ˆ
Ê3 3ˆ
7= 6+
1= 6¥ Á ˜ ¥ Á-
Ë
˜
(10 + 1) Á ¥
3 ¯ Ë 2x¯
Ë
˜
2 5¯
99 4
= = = 5
3 3 19 19
= 12 C 6
¥ x 6 Ê ˆ
= 1 + + ( ) + ( )
Thus, the co-efficient of x 10 is 30 C 5
.
20. We have,
= (1 + x) ¥ {(1 + x)(1 - x+
x
2 40
)}
(1 + x) 41 (1 – x + x 2 ) 40 3 40
= (1 + x) ¥ (1 + x )
40 3 40 3 2
= (1 + x) ¥ [1 + C1x + C1( x )
+
40
+ C
3 16 40
( x ) + C
3 17
( x ) + ]
Thus, the co-efficient of x 50 is 0.
21. We have,
(1 + x) 5 ¥ (1 – x) 6
Thus, co-efficient of x = 5 C 1
– 6 C 1
= 5 – 6 = –1
The co-efficient of x 2 = 6 C 2
+ 5 C 2
– 6 C 1
¥ 5 C 1
= 15 + 10 – 30 = –5
The co-efficient of x 3
= 5 C 3
– 6 C 3
+ 5 C 1
¥ 6 C 2
– 6 C 1
¥ 5 C 2
= 10 – 20 + 5.15 – 6.10
= 10 – 20 + 75 – 60
= 5
22. We have,
(1 + x) m (1 – x) n 2 2
= (1 + + + ) ¥ (1 - + - )
The co-efficient of x = 3
fi m C 1
– n C 1
= 3
fi m – n = 3 …(i)
The co-efficient of x 2 = –6
fi m C 2
+ n C 2
– m C 1
¥ m C 1
= –6
fi
mm
fi m(m – 1) + n(n – 1) – 2mn = –12
fi m 2 + n 2 – 2mn – (m + n) = –12
fi (m – n) 2 – (m + n) = –12
fi (–3) 2 – (m + n) = –12
fi –(m + n) = –21
fi (m + n) = 21 …(ii)
Solving Eqs (i) and (ii), we get
m = 12 and n = 9
Hence, the values of m and n are 12 and 9, respectively.
23. Since n is even, so the middle term = 12 1 7th
term.
Thus, t t C
25. Since n is even, so the middle term is (n + 1)th term.
Thus, t n+1
= 2n C n
¥ x n
26. The sum of the co-efficients of two middle terms in
(1 + x) 2n–1
= 2n – 1 C n – 1
+ 2n – 1 C n
= 2n – 1 C n
+ 2n – 1 C n – 1
= (2n–1)+1 C n
= 2n C n
= the co-efficient of the middle term in (1 + x) 2n .
27. Since n is even, so the greatest co-efficient
1 + Á ¥
Ë
˜
2 5¯
x
n