Question.1
The sum nC0
+nC1 +nC2+----+nCn
is eqaul to
(A) 
(B) nn
(C) n!
(D) 2n
Solution
(1
+ x)n = nC0 + nC1 x + nC2
x2 + nC3 x3 + L nCn xn
Put x = 1,
2n = nC0 + nC1
+ nC2 + L
+ nCn
Question.2
The digits at unit’s place
in the number 
is
(A) 0
(B) 1
(C) 2
(D) 3
Solution
We have
= 
= a multiple of 10 + 1
Thus, the unit’s place
digits is 1.
Question.3
The coefficient of x5
in the expansion of (1 + x)21 +(1 + x)22 + …..+ (1 + x)30
is.
(A) 51C5
(B) 9C5
(C) 31C6
-21C6
(D) 30C5
+ 20C5
Solution
(1 + x)21 +(1 + x)22 +
…..+(1 + x)30.
=
\ coefficient of x5
in the given expression
= coefficient of x5
in 
= coefficient of x6
in 
Question.4
If the co-efficients of x7 and x8
in 
are equal, then n is
(A) 56
(B) 55
(C) 45
(D) 15
Solution
&⇒
n = 55
Question.5
If z = 
, then
(A) Re(z) =0
(B) Im(z) =0
(C) Re(z) >0, Im(z)
>0
(D) Re(z) >0, Im(z)
<0
Solution
z = 2
= Purely real
number.
Hence Im(z) = 0
Question.6
If (1-x +x2)n
= a0+a1x+a2x2 +------+a2nx2n
then a0 +a2 +a4+---+a2n equals
(A) 
(B) 
(C) 
(D) 
Solution
Put
x = 1
&⇒ 1 = a0
+ a1 + a2 + L
+ a2n
Put x = -1
&⇒ 3n = a0
- a1 + a2
- a3 + L + a2n
Adding, 3n + 1 = 2 (a0 + a2
+ a4 + L+ a2n)
Question.7
The positive integer which is just greater
than (1+0.0001)1000 is .
(A) 3
(B) 4
(C) 5
(D) 2
Solution
Expression
on expansion gives
1 + 1000 ´
10-4 + 
< 1 + 
= 
So integer just greater than the given expression must
be 2.
Question.8
If n is an even natural number and coefficient of xr
in the expansion of 
is 2n, (|x|
< 1), then
(A) r £ n/2
(B) r ≥ 
(C) r £ 
(D) r ≥ n
Solution
= (C0 + C1x
+C2x2 + . . . + Cnxn) (1+ x+ x2
+ ….)
The coefficient of xr = C0 + C1+C2
+C3 + . . . + Cr = 2n for r = n.
Moreover coefficient of xr is C0 +
C1+C2 +C3 +. . .+ Cr if r > n.
So r ≥ n.
Question.9
The number of terms
in the expansion of (a+b+c)n, where n∈N, is
(A) 
(B) n+1
(C) n+2
(D) (n+1)n
Solution
(a
+ (b+c))n = an +nC1 an-1(b + c)1+nC2an-2 (b + c)2
+ L+ nCn
(b + c)n
Further
expanding each term of R.H.S.,.
First
term on expansion gives one term
Second
term on expansion gives two terms
Third
term on expansion gives three terms and so on.
\ Total no. of terms =
1 + 2 + 3 + L + (n + 1) = 
Question.10
If the co-efficient
of the second, third and fourth terms in the expansion of (1+x)n are
in A.P. then n is equal to .
(A) 2
(B) 7
(C) 9
(D) None of these
Solution
nC1,
nC2, nC3 are in A.P.
&⇒ nC1 + nC3
= 2nC2.
&⇒
n2 - 9n + 14 = 0
&⇒ n = 7,2
But n = 2 is rejected as nC3 is
not possible.
