1.Algebra Booster

Binomial Theorem 6.25
fi m π not an integer
Thus, the greatest term = 4
31. We have,
= 6 n .6 2 +7 2n .7
t È 5 ˘+
= 36.6 n + 7(6 + 43) n 1
Î 19˚
n n n n-1
= 36.6 + 7[6 + C
= t 6
= 6th term.
1◊ (6) 43 +
n n-2 2 n n-n n
C2
◊ + + Cn
◊
( n+
1)| x|
n n n-1 n n-1
m =
= 43[6 + C1
◊ (6) + + Cn
◊43 ]
a + | x|
Thus, it is divisible by 43.
Ê 5 1ˆ
16
(15 + 1) Á- ¥
36. We have,
Ë
˜
3 5¯
= =
3
11
= 4
10 – 1
Ê 5 1ˆ
4
10
1 + Á- ¥
= (1 + 10) -1
Ë
˜
3 5¯
3
10 10 2 10 3
= (1 + C1◊ 10 + C2◊ 10 + C3◊10
m = an integer
10 100
+ + C100
◊10 ) -1
10 10 2 10 3
= ( C1◊ 10+ C2◊ 10 + C3◊10
10 100
+ + C100
◊10 )
= (1 + 3) n – 3n – 1
10 10 10 98
= 100( C2 + C3 ◊ 10 + + C100
◊10 )
n n 2 n n
= (1 + C1◊ 3 + C2◊ 3 + + Cn
◊3 ) -3n-1
Thus, it is divisible by 100.
n 2 n n
= (1 + 3n+ C2
◊ 3 + + Cn
◊3 ) -3n-1
37. We have,
n 2 n 3 n n
7
= ( C2◊ 3 + C3◊ 3 + + Cn
◊3 )
98 = (7 2 ) 49
2 n n n n-
2
= 3( C2+ C3◊ 3+ + Cn
◊3 )
= (50 – 1) 49
49 49 48 49 47
= (50 - C1◊ 50 + C2◊50
49 46 49
- C3◊ 50 + - C49◊1)
49 49 48 49 47
= (50 - C1◊ 50 + C2◊50
49 46
- C
= 9.(1 + 8) n 3 ◊ 50 + - 5) + 4
– 8n – 9
Thus, the remainder is 4.
n n 2
= 9(1 + C1◊ 8 + C2◊8
38. We have
n 3 n n
+ C3
◊ 8 + + Cn
◊8 )-8n
-9
1 2013 + 2 2013 + 3 2013 + … + 2011 2013 + 2012 2013
n 2 n 3
= 9(8n + C2◊ 8 + C3◊8
n n
+ + Cn
◊8) -8n
2 n 2 n 3
= (8 n + C2◊8 ◊ 9 + C3◊8 ◊9
n n
+ + Cn
◊8 ◊9)
2 n n
= 8 ( n + C2◊ 9 + C3◊8.9
n n-
2
+ + Cn
◊8 ◊9)
= 121.11 n + 12.(11 + 133) n
n n n n-1
= 121.11 + 12[11 + C1
◊(11) 133
n n-2 2 n n-n n
+ C2
◊ (11) 133 + + Cn
◊(11) 133 ]
n n n-1 n n-1
= 133[11 + C1
◊ (11) + + Cn
◊133 ]
6 n+2 + 7 2n+1
fi
Thus, the greatest term are t 4
and t 4+1
, i.e 4th and 5th
terms.
32. We have,
(4 n – 3n – 1)
Thus, (4 n – 3n – 1) is divisible by 9.
33. We have,
3 2n+2 – 8n – 9
= 3 2 .3 2n – 8n – 9
= 9.9 n – 8n – 9
Thus, it is divisible by 64.
34. We have,
11 n+2 + 12 2n+1
= 11 2 .11 n + 12.12 2n
= 121.11 n + 12.(144) n
Thus, it is divisible by 133.
35. We have,
(6) 43 (6) 43 ]
= (1 2013 + 2012 2013 ) + (2 2013 + 2011 2013 )
+ … + (1006 2013 + 1007 2013 )
Here, each bracket is of the form of (a 2n+1 + b 2n+1 ) and so
is divisible by (a + b).
Hence, the given expression is divisible by 2013.
39. We have
1992 1998 – 1955 1998 – 1938 1998 + 1901 1998
= (1992 1998 – 1938 1998 ) – (1955 1998 – 1901 1998 )
Clearly, it is divisible by 54.
Also, the given expression can be written as
(1992 1998 – 1955 1998 ) – (1938 1998 – 1901 1998 )
It is divisible by 37.
Thus, the given expression
1992 1998 – 1955 1998 – 1938 1998 + 1901 1998
is divisible by 54 ¥ 37 = 1998
40. We have,
53 53 – 33 3
= ((53 53 – 43 53 ) + (43 53 – 33 3 ))
= ((53 53 – 43 53 ) + (43 53 – 33 43 ) + (33 43 – 33 3 ))
Clearly, each bracket is divisible by 10.