1.Algebra Booster

Binomial Theorem 6.31
n
n
n-1
n
n  r+ 1 Â
r-1
r
r= 0 r=
0
n
n
n-1
n
= nÂ
Cr-1
+ Â Cr
r= 0 r=
0
Thus, S = t = ( n◊ C + C )
= n.2 n–1 + 2 n
= (n + 2)2 n–1
82. Let t r
= r 2 .C r
=r 2 . n C r
n
= {( rr- 1) + r}
◊ Cr
n n
= {( rr- 1) Cr
+ rCr}
ÏÔ Ênn
( - 1) ˆ n-
2
= Ìrr
( - 1) ¥ Á Cr
2
rr ( 1) ˜ -
ÔÓ Ë - ¯
Ênˆ
n-1
¸
+ rÁ
Cr
-1˝
Ë
˜
r ¯ ˛
n-2 n-1
= {( nn- 1) C + n C }
Thus, S
n
n
n
= Â t
Â
r = 1
r
r = 1
r-2 r-1
n-2 n-1
r-2 r-1
= ( nn ( - 1) C + n C )
n
n
n-2 n-1
 r-2  r-1
r= 1 r=
1
= nn ( - 1) C + n C
= n(n – 1).2 n–2 + n.2 n–1
= n[(n – 1) ◊ 2 n – 2 + 2 n – 1 ]
= n ◊ 2 n – 2 (n – 1 + 2)
= n(n – 1)2 n–2
Cr
Cr
83. Let tr
+ 1 = =
r + 1 r + 1
1 Ên
+ 1 n ˆ
= ¥ Cr
n + 1Á
Ër
+ 1 ˜
¯
n + 1
Cr
+ 1
=
n + 1
n n Ê
n+
1
C ˆ
r + 1
Thus, Sn
= Âtr+
1 = ÂÁ ˜
Ë n + 1 ¯
n
r= 0 r=
0
n
1
n+
1
Â
= ¥ ( Cr
+ 1)
n + 1
r = 0
1 n+ 1 n+
1
= ¥ (2 - C0
)
n + 1
n+
1
2 - 1
=
n + 1
n
Cr
Cr
84. Let tr
+ 1 = ( r + 1)( r + 2) = ( r + 1)( r + 2)
Thus,
S
1 ( n + 1)( n + 2)
= ¥ ¥
( n + 1)( n + 2) ( r + 1)( r + 2)
n + 2
Cr
+ 2
=
( n + 1)( n + 2)
n
n
Â
= t =
n
Â
r+
1
r= 0 r=
0
1
=
( n + 1)( n + 2)
n + 2
Cr
+ 2
( n + 1)( n + 2)
n+
2
r = 0
n+ 2 n+ 2 n+
2
- C0-
C1
2
=
( n + 1)( n + 2)
n+
2
2 -1-( n + 2)
=
( n + 1)( n + 2)
n+
2
2 - n -3
=
( n + 1)( n + 2)
n
Â
( C )
r + 2
85. We have,
(1 + x) m = m C + m C ◊ x + m C
2
◊ x +
0 1 2
m r m r m m
Cr x Cr x Cm
x
2
0 1 2
n r n r + 1
C n n
r x Cr+
1 x Cn
x
+ ◊ + ◊ + + ◊
Also, (1 + x) n = n C + n C ◊ x + n C ◊ x +
n
C
r
…(i)
+ ◊ + ◊ + + ◊
…(ii)
Multiplying Eqs (i) and (ii), we get,
(1 + x) m+n m n m n
= ( Cr◊ C0+ Cr-1◊
C1
m n m n r
+ C ◊ C + + C ◊ C ) x
r-
2 2 0
+ (…)x + (…)x 2 + …
Comparing the co-efficients of x r from both the sides,
we get
m n m n m n
C ◊ C + C ◊ C + C ◊ C
r 0 r-1 1 r-2 2
m n m+
n
+ + C0 ◊ Cr)
= Cr
86. We have,
(1 + x) n = n C + n C ◊ x + n C
2
◊ x +
0 1 2
n r n r + 1
C n n
r x Cr+
1 x Cn
x
2
n 1 2
n r n r + 1
+ C n n
r ◊ x + Cr+
1 ◊ x + + Cn
◊ x
+ ◊ + ◊ + + ◊ …(i)
Also, (1 + x) n = n C x n + n C ◊ x+ n C ◊ x +
…(ii)
Multiplying Eqs (i) and (ii), we get
(1 + x) 2n n n n n
= ( C0◊ Cn
+ C1◊
Cn-1
n n n n n
+ C ◊ C + + C ◊ C ) x
2 n-
2 n 0
+ (…)x + (…)x 2 + …
Comparing the co-efficients of x n from both the sides,
we get
n n n n n n
( C ◊ C + C ◊ C + C ◊ C
0 n 1 n-1 2 n-2
n n 2n
+ + n ◊ 0 ) =
C C C
r
n