1.Algebra Booster

Binomial Theorem 6.23
11. Let t r+1
th term be the independent of x.
t
r+
1
6
2
6-
r
r
Ê3x
ˆ Ê 1 ˆ
= Cr
¥ Á ¥ -
Ë
˜
2 ¯ Á
Ë
˜
3x¯
6-
r r
6 Ê3ˆ Ê 1ˆ
12-2r-r
Cr
Á ˜ Á ˜ x
= ¥ ¥ - ¥
Ë2¯ Ë 3¯
Since the term is independent of x, so
12 – 3r = 0, i.e. r = 4
Thus, the 5th term is the independent of x.
12. Let t r+1
term has the co-efficient of x 7 .
Now,
11 2 11-
r Ê 1 ˆ
tr+
1 = Cr
¥ (3 x ) ¥Á
Ë
˜
5x¯
\ 22 – 3r = 7
fi 3r = 15
fi r = 5
r
11 11-r Ê1ˆ
22-2r-r
Cr
(3) Á ˜ x
= ¥ ¥ ¥
Ë5¯
Thus, the co-efficient of x 7 is
13. Given,
t 3
= 10000
5 5-
2 log
fi
10x
2
C x x
r
11 6
C5
3
2 ( ) ( ) = 10000
3 2log
fi
10x
x x
10 ◊ ( ) ( ) = 10000
3 2log
fi
10x
( x ) + = 1000
3 2log
fi
10x
3
( x ) + = 10
1
fi 3 + 2 log10x= log 3 x=
log
10
10x
3
Ê 1ˆ
fi Á2 - log10x
=-3
Ë
˜
3¯
9
fi log10x =-
5
fi x = 10 –9/5
14. We have, the co-efficient of x n in
(1 + x) 2n
= 2n C n
(2 n )!
=
( n)! ¥ ( n)!
2 n¥ (2n-1)!
=
( n)! ¥ ( n)!
2 ¥ (2n
-1)!
=
( n)! ¥ ( n-1)!
(2n
- 1)!
= 2 ¥
( n)! ¥ ( n-1)!
5
Ê1ˆ
¥ ¥Á
Ë
˜
5¯ .
= 2 ¥ 2n–1 C n
= 2 times the co-efficient of x n in (1 + x) 2n–1 .
15. Since,
2n 2n 2n
C , C , C ŒAP
fi
1 2 3
2 2 2
2 n C2= n C n
1+
C3
fi
fi
Ê2 n(2n-1) ˆ Ê2 n(2n-1)(2n-2)
ˆ
2◊ Á = 2n
+
Ë
˜ Á ˜
2 ¯ Ë 6 ¯
Ê2 n(2n-1)(2n-2)
ˆ
2 n(2n- 1) = 2n+Á Ë
˜
6 ¯
fi
(2n-1)(2n-2)
(2n
- 1) = 1+
6
fi (12n – 6) = 6 + (2n – 1)(2n – 2)
fi (12n – 6) = 6 + (4n 2 – 6n + 2)
fi (4n 2 – 18n + 14) = 0
fi (2n 2 – 9n + 7) = 0
16. Since,
n
C 4
, n C 5
, n C 6
are in AP
fi 2 n C 5
= n C 4
+ n C 6
fi 2 ! ! !
= n +
n
5! ¥ ( n- 5)! 4! ¥ ( n- 4)! 6! ¥ ( n-
6)!
fi
fi
fi
fi
2 1 1
= +
5 ¥ ( n-5) ( n- 4) ¥ ( n- 5) 6 ¥ 5
2 1 1
- =
5 ¥ ( n-5) ( n- 4) ¥ ( n- 5) 6 ¥ 5
2( n -4) -5 1
=
5 ¥ ( n- 4) ¥ ( n- 5) 6 ¥ 5
2n
- 13 1
=
( n- 4) ¥ ( n-5) 6
fi 6(2n – 13) = (n – 4) ¥ (n – 5)
fi 12n – 78 = n 2 – 9n + 20
fi n 2 – 21n + 98 = 0
fi (n – 7)(n – 14) = 0
fi n = 14, 7
Hence, the values of n are 7 and 14.
17 We have
t
n
= C x, t
n
= C x , t
n
= C x
2 3
2 1 3 2 4 3
n n n
Since C1, C2, C3
ŒAP
fi 2 n C 2
= n C 1
+ n C 3
fi
nn ( -1) nn ( -1)( n-2)
2 ◊ = n +
2 6
fi
nn ( -1)( n-2)
nn ( - 1) = n+
6
fi 6n(n – 1) = 6n + n(n – 1)(n – 2)
fi 6n 2 – 6n = 6n + n(n 2 – 3n + 2)
fi 6n 2 – 6n = 6n + n 3 – 3n 2 + 2n
fi n 3 – 9n 2 + 14n = 0
fi n(n 2 – 9n + 14) = 0
fi n(n – 2)(n – 7) = 0
fi n = 0, 2, 7
fi n = 7
18. The co-efficient of x 10 in (1 + x) 50 is 50 C 10
.
19. We have,
(1 + x 2 ) 30