1.Algebra Booster

Binomial Theorem 6.67
=2( n C 1
◊6 + n C 2
◊6 2 + n C 3
◊6 3 + … + n C n
◊6 n )
+ 3( n C 1
◊ 4 + n C 2
◊4 2 + … n C n
◊4 n )
Clearly, it is divisible by 24.
17. We have,
= n◊2 n–2 (n – 1 + 2)
= n◊(n + 1)2 n–2
(2 n)! 1
£
n
r r r
2
rn
È 1 3 7 15
˘
(2 n)! ¥ ( n) (3n
+ 1)
 (- 1) Cr Í + + + + m terms
r 2r 3r 4r
˙
r = 0 Î2 2 2 2
˚
20. Do yourself.
m n
k
n
n n
Ê2 - 1ˆ
2 1
= (-1) ( Cr
)
21. Let (5 5 11)
ÂÂ
n +
F = -
Á
Ë r ˜
k= 1r=
0
2 ¯
Then 0 < F < 1
m k
n
Also, R – F
Ê 2 - 1ˆ
2n+ 1 2n+
1
= Â Á1
-
= (5 5 + 11) - (5 5 -11)
Ë k ˜
k = 1 2 ¯
= Even integer
m n
Ê 1 ˆ
= 2m (say)
= Â Á
Ë k ˜
k = 1 2 ¯
Thus, R – F = 2m
fi [R] + f – F = 2m
m k
Ê 1 ˆ
fi f – F = 2m – [R]
= Â Á
Ë n ˜
k = 1 2 ¯
As, 0 < f < 1, 0 < F < 1
fi 0 < f < 1, –1 < –F < 0
2 3
m
1 Ê 1 ˆ Ê 1 ˆ Ê 1 ˆ
fi –1 < f – F < 1
= + + + +
n Á n˜ Á n˜ Á n˜
2 Ë2 ¯ Ë2 ¯ Ë2
¯
fi f – F = 0
fi f = F
1 Ê 1 ˆ
1 -
n Á mn˜
mn
Now,
2 Ë 2 ¯ 2 - 1
2n+ 1 2n+
1
= =
Rf = (5 5 + 11) (5 5 -11)
1
mn n
1 -
2 (2 - 1)
2n+
1
n
2
= ((5 5 + 11) ¥ (5 5 -11))
=(125 – 121) 2n+1
= 4 2n+1
Ênˆ Ênˆ
22. We have,
2 Á ˜! Á ˜!
Ë2¯ Ë2¯ ¥
2 2 n 2
[ C0 + C1
+ + ( - 1) ( n + 1) Cn
]
n!
where n is an even +ve integer, is equal to
Also,
n + 1
(a) 0
(b) (-1)
2
(c) (–1) n (n + 2) (d) None
[IIT-JEE, 1986]
= [k(k – 1)] n C k
+ k n C k
nn ( - 1) n-2 n n-1
= kk ( - 1) ¥ Ck-2+ k¥
Ck-1
kk ( - 1)
k
= n(n – 1) n–2 C k–2
+ n n–1 C k–1
= m+n C k
Therefore,
n
n
2 n
Âk Ck
= Âtk+
1
k= 0 k=
0
n
n-2 n-1
= Â ( nn ( - 1) Ck-2+
n Ck-1)
k = 0
n
n
n-2 n-1
= nn ( - 1) Â( Ck-2) + nÂ( Ck-1)
k= 0 k=
0
= n(n – 1)2 n–2 + n2 n–1
18. If C r
stands for n C r
, then sum of the series
19. We have
t k+1
= k 2 C k
= [k(k – 1) + k] n C k
20. Prove by the mathematical induction that
for every +ve integer n.
(1 + x) m
= m C 0
+ m C 1
x + m C 2
x 2 + … + m C k
x k + …
[IIT-JEE, 1987]
…(i)
(1 + x) n
= n C 0
+ n C 1
x + n C 2
x 2 + n C 3
xx + … + n C k
x k + n C k–1
x k–1
+ n C k–2
x k–2 + …
…(ii)
Multiplying Eqs (i) and (ii), we get
(1 + x) m+n = ( m C 0
◊ n
C k
+ m C 1
◊ n
C k–1
+ m C 2
◊ n
C k–2
+ … + m C k
◊ n C 0
)x k + (…)x k–1 + (…)x k–2 + …
Comparing the co-efficients of x k from both the sides,
we get
m
C 0
◊ n C k
+ m C 1
◊ n C k–1
+ m C 2
◊ n C k–2
+ … + m C k
◊ n C 0
23. We have,
C 0
+ C 1
x + C 2
x 2 + … + C n
x n = (1 + x) n
Multiplying by x, we get
C 0
x + C 1
x 2 + C 2
x 3 + … + C n
x n+1 = x(1 + x) n
Differentiating w.r.t x, we get
C 0
+ 2 2 C 1
x + 3 2 C 2
x 2 + … + (n + 1)C n
X n
= (1 + x) n + nx(1 + x) n–1
Again multiplying by x, we get
C 0
x + 2C 1
x 2 + 3C 2
x 3 + … + (n + 1)C n
x n+1
= x(1 + x) n + nx 2 (1 + x) n–1