1.Algebra Booster

6.68 Algebra Booster
Differentiating w.r.t x, we get
C 0
+ 2 2 C 1
x + 3 2 C 2
x 2 + … + (n + 1) 2 C n
x n
= nx(1 + x) n–1 + n(n – 1)x 2 (1 + x) n–2 + 2nx(1 + x) n–1
Putting x = –1, we get
C 0
– 2 2 C 1
+ 3 2 C 2
+ … + (–1) n (n + 1) 2 C n
= 0
24. Do yourself, by mathematical induction.
25. The product of n +ve numbers is unity.
Then their sum is
(a) a +ve integer (b) divisible by n
1
(c) equal to n + (d) never less than n.
n
[IIT-JEE, 1991]
26. We have,
k
Ê( r + m)!
ˆ
 ( n - m)
Á
Ë k!
˜
¯
m=
0
k
Ê Ê( r + m)! ˆ Ê( r + m)!
ˆˆ
= Â Á
n - m
Ë
Á
Ë k! ˜
¯
Á
Ë k!
˜
¯˜
¯
m=
0
k
Ê Ê( r + m)! ˆ Ê( r + m)!
ˆˆ
= ( r!
) Â Á
n - m
m=
0Ë
Á
Ë r! ¥ k! ˜
¯
Á
Ë r! ¥ k!
˜
¯˜
¯
Ê
k
k
r+ m r+
m
ˆ
= ( r!)
ÁnÂ
Cm
- mÂ
Cm˜
Ë m= 0 m=
0 ¯
= ( r!) È
Î
n[ C + C + C + + C ]
r r+ 1 r+ 2
r+
k
0 1 2
k
k
r+
m
˘
- Â [( m + r + 1) - ( r + 1) Cm
]
r = 0
˙˙˚
r+ k+ 1 r+ k+ 2 r+ k+
1
k k k
r+ k+ 1 r+ k+
1
k
k-1
= r![ n( C ) -( r + 1)( C ) -( C )]
= r![ n( C ) -( r + 1)( C )]
È Ê ( r + k + 1)! ˆ Ê ( r + k + 1)! ˆ˘
= r! ÍnÁ ( r 1)
( r 1)! k! ˜ - + Á( r 2)! ( k 1)! ˜˙
Î Ë + ¥ ¯ Ë + ¥ - ¯˚
È( r + k + 1)! ˘È n k ˘
= Í k! ˙Í -
r + 1 r + 2˙
Î ˚Î ˚
Hence, the result.
27. Given, (a 2 x 2 – 2ax + 1) 51 = 0
Putting x = 1, we get
(a 2 – 2a + 1) 51 = 0
fi (a 2 – 2a + 1) = 0
fi (a – 1) 2 = 0
fi (a – 1) = 0
fi a = 1
28. We have,
fi
2n
2n
r
Âar
x Âbr
x
r= 0 r=
0
2n
2n
r
 r  r
r= 0 r=
0
( - 2) = ( -3)
a (( x - 3) + 1) = b ( x -3)
r
r
fi
2n
2n
r
 r  r
r= 0 r=
0
a ( y + 1) = b y , y = ( x -3)
Comparing the co-efficients from both the sides, we get
2n
b = a ( C )
n
=
Â
r
r = 0
2n
r
Â
r=
n
r
( C )
n
n
n n+ 1 n+
2 2n
| Cn Cn Cn Cn|
n n+ 1 n+
2 2n
| C0 C1 C2
Cn|
2n+
1
Cn
2n+
1
Cn+
1
= + + + +
= + + + +
=
=
29. We have,
100
100 100 - m m
 Cm
( x - 3) 2 = ((x – 3) + 2) 100
m=
0
= (x – 1) 100
= (1 – x) 100
\ Co-efficient of x 53 = (–1) 53 100 C 53
= – 100 C 53
.
30. We have,
( x +
3 5
x - 1) + ( x -
3 5
x -1)
= (x + a) 5 + (x – a) 5 , where a = x -1
= 2(x 5 + 5 C 2
x 3 a 2 + 5 C 4
xa 4 )
= 2[x 5 + 5 C 2
x 3 (x 3 – 1) + 5 C 4
x(x 3 – 1) 2 ]
Thus, the degree of the polynomial is 7.
31. Let t r+1
= (2r + 1)C r
= (2r + 1) n C r
= 2r n C r
+ n C r
n n-1
n
= 2r ¥ ¥ Cr-1
+ Cr
r
= 2n ¥ n–1 C r–1
+ n C r
Thus, S = t + 1
n
n
Â
r
r = 0
n
n-1
n
 n Cr-1
Cr
r = 0
n
n
n-1
n
2 Â r-1
Â
r= 0 r=
0
= (2 + )
= n C + C
= 2n◊2 n–1 + 2 n
= n.2 n + 2 n
= (n + 1)2 n
32. Let n = 2 m, so that k = 3 m
We have
k
 (-3)
r = 1
3m
Â
r = 1
r-13n
= (-3)
C
2r
-1
r-16m
C
r
2r
-1
r
3