1.Algebra Booster

Binomial Theorem 6.43
and
D
Now, n + 1 C 1
+ n + 1 C 2
S 1
+ n + 1 C 3
S 3
+ … + n + 1 C n + 1
S n
25. Given, a = (4 1/401 – 1)
fi (a + 1) 401 = 4
Now, b n
= n C 1
+ n C 2
◊ a + n C 3
◊ a 2 + … + n C n
◊ a n – 1
\ 1 + ab n
= n C 1
◊ a + n C 2
◊ a 2 + n C 3
◊ a 3 + … + n C n
◊ a n
fi
fi
n+
1
1 - q
fi
=
1 - q
2
n
Êq + 1ˆ Êq + 1ˆ Êq
+ 1ˆ
fi
n = 1 + Á + + +
Ë
˜ Á ˜ Á ˜
2 ¯ Ë 2 ¯ Ë 2 ¯
fi
n+
1
Ê1
+ qˆ
1 - Á n+ 1 n+
1
Ë
˜
2 ¯ 2 - (1 + q)
= =
1 q
n
Ê + ˆ
1 -
2 (1-
q)
Á
Ë
˜
2 ¯
1 [
n+
1
= C1
(1 - q )
1 - q
n+ 1 2 n+
1 3
+ C2(1 - q ) + C3(1 - q )
n+ 1 n+
1
+ + Cn+
1(1 - q )]
n 1 n 1
1 È
+ +
˘
n+ 1 n+
1 k
= ÍÂ
Ck
-Â
Ckq
˙
1 - q Í Îk= 1 k=
1 ˙ ˚
1 ((2
n+ 1 1) {(1 )
n+
1
= - - + q - 1})
1 - q
n+ 1 n+
1
2 - (1 + q)
=
(1 - q)
n
= 2 Dn
fi
27. Let
fi
1 + ab n
= (1 + a) n
n
(1 + a) -1
bn
=
a
2006 2005
(1 + a) - 1 (1 + a) -1
= -
a
a
2006 2005
(1 + a) -(1 + a)
=
a
2005
(1 + a) (1 + a -1)
=
a
2005
( 1 + a)
¥ a
=
a
= (1 + a) 2005
fi
– n C 4
a n – 4 + … + (–1) n – 1 n C n – 1
Hence,
the value of (b 2006
– b 2005
)
2005
= 4 401
= 4 5 = 2 10
26. We have,
f(n) = n C 0
◊ a n – 1 – n C 1
◊ a n – 2 + n C 2
◊ a n – 3
n n n n n-1 n n-2
0 1 2
n n-3 n-1
n
Ca 4 ( 1) Cn-1
a 1
af ( n) + (- 1) = C ◊a - C ◊ a + C ◊a
- + + - ◊ -
af(n) + (–1) n = (a – 1) n
( a -1) - (-1)
f( n)
=
a
Now,
f(2007) + f(2008)
n
n
2007 2007 2008 2008
( a -1) - (-1) ( a -1) - (-1)
= +
a
a
2007 2008
( a - 1) + 1 ( a -1) -1
= +
a
a
2007 2008
( a - 1) + ( a -1)
=
a
2007
( a - 1) (1+ a -1)
=
a
2007
( a - 1) ¥ a
=
a
2007
= ( a -1)
2007
323
= 3 = 3
Now, 9m = 3 9
9
3 7
m = = 3 = 2187
9
n
Â
k = 0
n
9
S = C ◊sin( kx) ◊cos( n - k)
x
n
Â
k = 0
n
k
S = C ◊sin[( n - k) x] ◊cos( k)
x
Adding, we get
n
Â
n
k = 0
n
k = 0
n-
k
2S = C ◊sin( kx) ◊cos( n - k)
x
n
Â
k = 0
n
n
Â
n
k
+ C◊sin[( n- kx ) ] ◊cos( kx )
n
k
= C ◊ sin[ kx + ( n - k) x]
Â
k = 0
k
= C ◊sin( n)
x
k
n
= sin( nx ) Â
k = 0
= sin(n)x ¥ 2 n
S = 2 n – 1 ¥ sin(n)x
28. Let d be the common difference.
So, a n–1
= a + nd
We have,
n
Â
k = 0
n
Ck
◊ a k + 1
n
C
k