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CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 7 - MCQExams.com
CBSE
Class 11 Engineering Maths
Binomial Theorem
Quiz 7
The value of
C
2
1
+
C
2
2
.
.
.
.
+
C
2
n
(where
C
i
is the
i
t
h
coefficient of
(
1
+
x
)
n
expansion), is:
Report Question
0%
n
n
n
!
0%
2
n
!
n
!
n
!
0%
2
n
!
n
!
0%
n
!
×
2
n
2
n
!
Explanation
(
1
+
x
)
n
=
n
C
0
+
x
(
n
C
1
)
+
x
2
(
n
C
2
)
+
x
3
(
n
C
3
)
+
.
.
.
+
x
n
(
n
C
n
)
(
x
+
1
)
n
=
x
n
(
n
C
0
)
+
x
n
−
1
(
n
C
1
)
+
.
.
.
+
x
0
(
n
C
n
)
Now,
n
∑
r
=
0
(
n
C
r
)
2
= co-efficient of
x
n
in
(
1
+
x
)
2
n
=
2
n
C
n
Thus,
2
n
C
n
=
(
n
C
0
)
2
+
(
n
C
1
)
2
+
.
.
.
(
n
C
n
)
2
, where
(
n
C
i
)
is the
i
t
h
coefficient of
(
1
+
x
)
n
expansion.
So,
C
2
1
+
C
2
2
+
.
.
.
+
C
2
n
=
(
2
n
C
n
)
=
2
n
!
n
!
n
!
If
f
(
n
)
=
n
∑
s
=
1
n
∑
r
=
s
n
C
r
r
C
s
, then
f
(
3
)
=
Report Question
0%
27
0%
19
0%
1
0%
5
Explanation
Given :
f
(
n
)
=
n
Σ
s
=
1
n
Σ
r
=
s
n
C
r
r
C
s
To Find :
f
(
3
)
=
?
Sol. :
f
(
n
)
=
{
n
C
1
1
C
1
+
n
C
2
2
C
1
+
n
C
3
3
C
1
.
.
.
.
.
.
n
C
n
.
n
C
1
}
+
{
n
C
2
2
C
2
+
n
C
3
3
C
2
+
.
.
.
.
.
n
C
n
n
C
2
}
+
.
.
.
.
.
+
{
n
C
n
−
1
n
−
1
C
n
−
1
+
n
C
n
n
C
n
−
1
}
+
{
n
C
n
n
C
n
}
⇒
Take
n
=
3
⇒
f
(
3
)
=
{
3
C
1
1
C
1
+
3
C
2
2
C
1
+
3
C
3
3
C
1
}
+
{
3
C
2
2
C
2
+
3
C
3
3
C
2
}
+
{
3
C
3
3
C
3
}
=
{
3
+
3.2
+
3
}
+
{
3
+
3
}
+
{
1
}
=
19
Hence, the answer is
19.
The value of
(
n
+
2
)
C
0
2
n
+
1
−
(
n
+
1
)
C
1
2
n
+
n
C
2
2
n
−
1
+
.
.
.
.
is equal to:
(
C
r
=
n
C
r
)
Report Question
0%
4
n
0%
4
0%
2
n
+
4
0%
4
+
4
n
Explanation
We know,
(
1
−
x
)
n
=
n
C
0
x
n
−
n
C
1
x
n
−
1
+
n
C
2
x
n
−
2
−
.
.
.
.
n
C
n
Multiplying by
x
2
on both sides,
x
2
(
1
−
x
)
n
=
n
C
0
x
n
+
2
−
n
C
1
x
n
+
1
+
n
C
2
x
n
−
.
.
.
.
Taking the derivative,
2
x
(
1
−
x
)
n
+
n
x
2
(
1
−
x
)
n
−
1
=
(
n
+
2
)
C
0
x
n
+
1
−
(
n
+
1
)
C
1
x
n
+
n
C
2
x
n
−
1
−
.
.
.
.
Putting
x
=
2
in LHS, we get our required result,
4
(
−
1
)
n
+
n
(
4
)
(
−
1
)
n
−
1
=
4
−
4
n
,
n
is even
=
4
n
−
4
,
n
is odd.
There is
n
o
o
p
t
i
o
n
matching the answer.
If
n
−
1
C
r
=
(
k
2
−
3
)
n
C
r
+
1
, then
k
ϵ
Report Question
0%
(
−
∞
,
−
2
]
0%
[
2
,
∞
)
0%
[
−
√
3
,
√
3
]
0%
(
√
3
,
2
]
Explanation
Given,
n
−
1
C
r
=
(
k
2
−
3
)
n
C
r
+
1
⇒
n
−
1
C
r
n
C
r
+
1
=
(
k
2
−
3
)
....Using
n
C
x
=
n
x
.
n
−
1
C
x
−
1
⇒
r
+
1
n
=
(
k
2
−
3
)
....As
n
>
r
+
1
So,
0
<
r
+
1
n
≤
1
Therefore,
0
<
k
2
−
3
≤
1
⇒
3
<
k
2
≤
4
⇒
√
3
<
k
≤
2
So, k
∈
(
√
3
,
2
]
If
P
n
denotes the product of all the coefficients in the expansion of
(
1
+
x
)
n
, then
P
n
+
1
P
n
is equal to:
Report Question
0%
(
n
+
2
)
n
n
!
0%
(
n
+
1
)
n
+
1
n
+
1
!
0%
(
n
+
1
)
n
+
1
n
!
0%
(
n
+
1
)
n
n
+
1
!
Explanation
(
1
+
x
)
n
P
n
+
1
=
n
+
1
C
0
×
n
+
1
C
1
×
n
+
1
C
2
×
.
.
.
.
.
.
.
.
×
n
+
1
C
n
+
1
P
n
=
n
C
0
×
n
C
1
×
n
C
2
×
.
.
.
.
.
.
.
.
×
n
C
n
n
∏
r
=
0
n
+
1
C
r
n
C
r
(
∵
n
+
1
C
n
+
1
=
1
)
=
n
∏
r
=
0
(
n
+
1
)
!
r
!
(
n
+
1
−
r
)
!
×
r
!
(
n
−
r
)
!
n
!
=
n
∏
r
=
0
(
n
+
1
)
(
n
−
r
+
1
)
=
(
n
+
1
)
n
n
∏
r
=
0
1
n
−
r
+
1
=
(
n
+
1
)
n
(
n
+
1
)
!
Hence, the correct option is
D
.
The value of
50
C
4
+
6
∑
r
=
1
56
−
r
C
3
is
Report Question
0%
55
C
4
0%
55
C
3
0%
56
C
3
0%
56
C
4
Explanation
50
C
4
+
6
∑
r
=
1
56
−
r
C
3
50
C
4
+
50
C
3
+
51
C
3
+.................+
55
C
3
Using
n
C
r
+
n
C
r
−
1
=
n
+
1
C
r
51
C
4
+
51
C
3
+.................+
55
C
3
Similarly the series would become like this
55
C
4
+
55
C
3
Using
n
C
r
+
n
C
r
−
1
=
n
+
1
C
r
56
C
4
The coefficient of
x
53
in the following expansions.
100
∑
m
=
0
100
C
m
(
x
−
3
)
100
−
m
⋅
2
m
is
Report Question
0%
100
C
47
0%
100
C
53
0%
−
100
C
53
0%
−
100
C
100
Explanation
100
∑
m
=
0
100
C
m
(
x
−
3
)
100
−
m
⋅
2
m
Above expansion can be rewritten as
[
(
x
−
3
)
+
2
]
100
=
(
x
−
1
)
100
=
(
1
−
x
)
100
∴
x
53
will occur in
T
54
i.e.
54
t
h
term.
So,
m
=
53
⇒
T
54
=
100
C
53
(
−
x
)
53
∴
Required coefficient is
−
100
C
53
.
The coefficient of
x
2012
in
1
+
x
(
1
+
x
2
)
(
1
−
x
)
is.
Report Question
0%
1
0%
2
0%
3
0%
4
Explanation
(
1
+
x
)
(
1
+
x
2
)
(
x
−
1
)
=
x
(
1
+
x
2
)
+
1
(
1
−
x
)
The general term for
(
1
−
x
)
−
1
=
1
+
x
+
x
2
+
x
3
+
.
.
.
.
.
.
.
.
.
(
1
−
x
)
−
1
=
∞
∑
0
x
n
.
.
.
.
.
.
.
.
.
(
i
)
Replacing
x
by
−
x
2
1
1
+
x
2
=
1
−
x
2
+
x
4
−
x
6
.
.
.
.
.
.
.
1
1
+
x
2
=
∞
∑
0
(
−
1
)
n
x
2
n
Multiplying both sides by
x
x
1
+
x
2
=
∞
∑
0
(
−
1
)
n
x
2
n
+
1
.
.
.
.
(
i
i
)
From (i) clearly the cofficient of
x
2012
is
1
and there is no term with even cofficient in
(
i
i
)
So the cofficient of
x
2012
in the given expression is
1
Let
n
be a positive integer and,
(
1
+
x
)
n
=
a
0
+
a
1
x
+
a
2
x
2
+
⋯
+
a
n
x
n
. What is
a
0
+
a
1
+
a
2
+
⋯
+
a
n
equal to?
Report Question
0%
1
0%
2
n
0%
2
n
−
1
0%
2
n
+
1
Explanation
Given expansion is
(
1
+
x
)
n
=
a
0
+
a
1
x
+
a
2
x
2
+
.
.
.
.
a
n
x
n
Substituting
x
=
1
on both the sides we get,
(
1
+
1
)
n
=
a
0
+
a
1
×
1
+
a
2
×
1
2
+
.
.
.
.
a
n
×
1
n
⟹
2
n
=
a
0
+
a
1
+
a
2
+
.
.
.
a
n
Thus, the value of
a
0
+
a
1
+
a
2
+
.
.
.
a
n
is
2
n
.
5
t
h
term from the end in the expansion of
(
x
2
2
−
2
x
2
)
12
is
Report Question
0%
−
7920
x
−
4
0%
7920
x
4
0%
7920
x
−
4
0%
−
7920
x
4
Explanation
5
t
h
term in the expansion of
(
x
2
2
−
2
x
2
)
12
is
T
r
+
1
=
n
C
r
x
r
.
y
n
−
r
T
5
=
12
C
4
[
(
x
2
2
)
4
(
−
2
x
2
)
8
]
=
12
!
4
!
8
!
.
x
8
2
4
.
2
8
x
16
=
7920
x
−
4
Consider the expansion of
(
1
+
x
)
2
n
+
1
The average of the coefficients of the two middle terms in the expansion is
Report Question
0%
2
n
+
1
C
n
+
2
0%
2
n
+
1
C
n
0%
2
n
+
1
C
n
−
1
0%
2
n
C
n
+
1
Explanation
Since,
2
n
+
1
is odd.
Hence,
2
n
+
1
+
1
2
and
2
n
+
1
+
3
2
are two middle terms.
i.e.
(
n
+
1
)
t
h
and
(
n
+
2
)
t
h
terms are two middle terms.
∴
C
2
n
+
1
n
+
C
2
n
+
1
n
+
1
2
=
C
2
n
+
1
+
1
n
+
1
2
=
1
2
C
2
n
+
2
n
+
1
=
1
2
.
2
n
+
2
n
+
1
C
2
n
+
1
n
=
C
2
n
+
1
n
Hence, B is the correct option.
In the expansion of
(
x
3
−
1
x
2
)
n
,
n
∈
N
, if the sum of the coefficient of
x
5
and
x
10
is
0
, then
n
is :
Report Question
0%
25
0%
20
0%
15
0%
None of these
Explanation
Term of
x
5
=
n
C
r
x
3
r
(
−
1
x
2
)
n
−
r
=
n
C
r
x
3
r
.
x
2
r
−
2
n
.
(
−
1
)
n
−
r
So
3
r
+
2
r
−
2
n
=
5
r
=
5
+
2
n
5
⟶
(
1
)
Also for
x
10
be
(
r
1
+
1
)
t
h
term
So term with
x
10
=
n
C
r
1
x
3
r
1
(
−
1
x
2
)
n
−
r
⇒
3
r
1
+
2
r
1
−
2
n
=
10
r
1
=
2
n
+
10
5
⟶
(
2
)
Now we know if
|
n
C
r
|
=
|
n
C
r
1
|
⇒
n
=
r
+
r
1
Now we add co oefficient of
x
5
&
x
10
n
C
r
(
−
1
)
n
−
r
+
n
C
r
1
(
−
1
)
n
−
r
1
=
0
Co oefficient of
x
5
i
s
n
C
r
(
−
1
)
n
−
r
; co oeffiecient of
x
10
i
s
n
C
r
1
(
−
1
)
n
−
r
1
⇒
n
C
r
(
−
1
)
n
−
r
=
−
(
n
C
r
1
(
−
1
)
n
−
r
1
)
⇒
|
n
C
r
|
=
|
n
C
r
1
|
So,
n
=
r
+
r
1
n
=
2
n
+
10
5
+
2
n
+
5
5
5
n
=
4
n
+
15
n
=
15
The value of
n
C
0
−
n
C
1
+
n
C
2
−
.
.
.
.
+
(
−
1
)
n
n
C
n
is:
Report Question
0%
1
0%
0
0%
2
n
0%
n
Explanation
By Binomial Expansion we know,
(
a
+
b
)
n
=
n
C
0
a
0
b
n
+
n
C
1
a
1
b
n
−
1
+
.
.
.
.
+
n
C
n
−
1
a
n
−
1
b
1
+
n
C
n
a
n
b
0
Now putting
a
=
−
1
&
b
=
+
1
we have,
(
−
1
+
1
)
n
=
n
C
0
(
−
1
)
0
(
+
1
)
n
−
1
+
n
C
0
(
−
1
)
1
(
+
1
)
n
−
1
+
.
.
.
.
.
+
n
C
n
−
1
(
−
1
)
n
−
1
(
+
1
)
+
n
C
n
(
−
1
)
n
(
+
1
)
0
⇒
0
=
n
C
0
(
−
1
)
0
(
+
1
)
n
−
1
+
n
C
0
(
−
1
)
1
(
+
1
)
n
−
1
+
.
.
.
.
.
+
n
C
n
−
1
(
−
1
)
n
−
1
(
+
1
)
+
n
C
n
(
−
1
)
n
The value of the expression
k
−
1
C
k
−
1
+
k
C
k
−
1
+
.
.
.
.
n
+
k
−
2
C
k
−
1
is given by :
Report Question
0%
n
+
k
−
1
C
k
−
1
0%
n
+
k
−
1
C
k
0%
n
+
k
C
k
0%
None of these
Explanation
We know,
n
∑
m
=
k
m
C
r
=
n
+
1
C
r
+
1
According to the question,
m
=
n
+
k
−
2
r
=
k
−
1
Sum of series is given by
=
m
+
1
C
k
+
1
=
n
+
k
−
1
C
k
How many terms are there in the expansion of
(
1
+
2
x
+
x
2
)
10
?
Report Question
0%
11
0%
20
0%
21
0%
30
Explanation
Now,
(
1
+
2
x
+
x
2
)
10
=
(
(
1
+
x
)
2
)
10
=
(
1
+
x
)
20
Now, the number of terms in the expansion of
(
1
+
x
)
n
are
n
+
1
.
Thus, the number of terms in the expansion of
(
1
+
x
)
20
will be
20
+
1
=
21
.
Hence, option C is correct.
Consider the expansion of
(
1
+
x
)
2
n
+
1
The sum of the coefficients of all the terms in the expansion is
Report Question
0%
2
2
n
−
1
0%
4
n
−
1
0%
2
×
4
n
0%
None of the above
Explanation
We can get the sum of all coefficients by putting
x
=
1
in the expansion, because for calculating the coefficients we need the terms independent of
x
.
To find the sum of coefficients of all terms, put
x
=
1
in the given expression
(
1
+
x
)
2
n
+
1
we get
2
2
n
+
1
=
2.2
2
n
=
2.4
n
Hence, C is the correct option.
What is n equal to ?
Report Question
0%
5
0%
10
0%
15
0%
None of the above
Explanation
Using binomial theorem, we can write
(
x
3
−
1
x
2
)
n
=
n
∑
r
=
0
n
r
C
(
x
3
)
n
−
r
(
−
1
x
2
)
r
=
n
∑
r
=
0
(
−
1
)
r
n
r
C
x
(
3
n
−
5
r
)
For
x
3
and
x
10
let value of r be
r
5
and
r
10
respectively
3
n
−
5
r
10
=
5
.........
(
i
)
3
n
−
5
r
10
=
10
..........
(
i
i
)
(
−
1
)
r
5
n
r
5
C
x
(
3
n
−
5
r
5
)
=
−
(
−
1
)
r
10
n
r
10
C
x
(
3
n
−
5
r
10
)
Given sum of coefficient of
x
5
and
x
10
is 0
All terms are positive, but still there is a negative sign, means one of
r
5
and
r
10
is even and other is odd
By properties of combinations
n
r
5
C
=
n
r
10
C
r
5
+
r
10
=
n
..........
(
i
i
i
)
Adding
(
i
)
and
(
i
i
)
6
n
−
5
(
r
5
+
r
10
)
=
15
From
(
i
i
i
)
,
r
5
+
r
10
=
n
⇒
n
=
15
In the expansion of
(
x
+
1
x
)
n
, then the coefficient of the term indepenent of x is
Report Question
0%
n
!
(
r
!
)
2
0%
n
!
(
r
+
1
)
!
(
r
−
1
)
!
0%
n
!
(
n
+
r
2
)
!
(
n
−
r
2
)
!
0%
n
!
[
(
n
2
)
!
]
2
Explanation
In the expansion of
(
x
+
1
x
)
n
,
r
t
h
term is
t
r
+
1
=
n
C
r
x
n
−
r
(
1
x
)
r
⟹
t
r
+
1
=
n
C
r
x
n
−
r
x
−
r
⟹
t
r
+
1
=
n
C
r
x
n
−
2
r
...
(
i
)
We need to find the coefficient of
x
0
Then substitute
n
−
2
r
=
0
⟹
r
=
n
2
Substitute
r
in RHS of
(
i
)
, we get
t
r
+
1
=
n
C
n
2
x
Then, coefficient of
x
is
=
n
C
n
2
=
n
!
(
n
2
)
!
(
n
−
n
2
)
!
=
n
!
(
n
2
)
!
(
n
2
)
!
=
n
!
[
(
n
2
)
!
]
2
Hence, coefficient of
x
0
=
x
is
n
!
[
(
n
2
)
!
]
2
The sum of the series
20
C
0
−
20
C
1
+
20
C
2
−
20
C
3
+
.
.
.
+
20
C
10
is
Report Question
0%
−
20
C
10
0%
1
2
20
C
10
0%
0
0%
20
C
10
Explanation
On putting
x
=
−
1
in
(
1
+
x
)
20
=
20
C
0
+
20
C
1
x
+
.
.
.
+
20
C
10
x
10
+
.
.
.
+
20
C
20
x
20
We get,
0
=
20
C
0
−
20
C
1
+
.
.
.
−
20
C
9
+
20
C
10
−
20
C
11
+
.
.
.
+
20
C
20
⇒
0
=
20
C
0
−
20
C
1
+
.
.
.
−
20
C
9
+
20
C
10
−
20
C
9
+
.
.
.
+
20
C
0
........
[
∵
n
C
r
=
n
C
n
−
r
]
⇒
0
=
2
(
20
C
0
−
20
C
1
+
.
.
.
−
20
C
9
)
+
20
C
10
⇒
20
C
10
=
2
(
20
C
0
−
20
C
1
+
.
.
.
+
20
C
10
)
.... [Adding
20
C
10
on both the sides]
⇒
20
C
0
−
20
C
1
+
.
.
.
+
20
C
10
=
1
2
20
C
10
If
T
r
=
2016
C
r
x
2016
−
r
, for
r
=
0
,
1
,
,
.
.
.
.2016
, then
(
T
0
−
T
2
+
T
4
.
.
.
.
+
T
2016
)
2
+
(
T
1
−
T
3
+
T
5
.
.
.
.
T
2015
)
2
is equal to -
Report Question
0%
(
X
2
−
1
)
1008
0%
(
X
+
1
)
2016
0%
(
X
2
−
1
)
2016
0%
(
X
2
+
1
)
2016
Explanation
Let
i
represent iota.
(
T
0
−
T
2
+
T
4
+
⋯
+
T
2016
)
=
(
T
0
+
i
2
T
2
+
i
4
T
4
+
⋯
+
i
2016
T
2016
)
…
(
1
)
Now, multiply
(
T
1
−
T
3
+
T
5
+
⋯
−
T
2015
)
with
−
i
to get
−
(
i
T
1
+
i
3
T
3
+
i
3
T
5
+
⋯
+
i
2015
T
2015
)
…
(
2
)
⟹
(
T
0
−
T
2
+
T
4
+
⋯
+
T
2016
)
2
+
(
T
1
−
T
3
+
T
5
+
⋯
−
T
2015
)
2
=
(
T
0
+
i
2
T
2
+
i
4
T
4
+
⋯
+
i
2016
T
2016
)
2
+
i
4
(
T
1
−
T
3
+
T
5
+
⋯
−
T
2015
)
2
=
(
T
0
+
i
2
T
2
+
i
4
T
4
+
⋯
+
i
2016
T
2016
)
2
−
(
i
T
1
+
i
3
T
3
+
i
3
T
5
+
⋯
+
i
2015
T
2015
)
2
=
(
T
0
+
i
T
1
+
i
2
T
2
+
⋯
+
i
2016
T
2016
)
(
T
0
−
i
T
1
+
i
2
T
2
−
i
3
T
3
…
)
(As
a
2
−
b
2
=
(
a
+
b
)
(
a
−
b
)
)
Now, consider these terms seperately
(
T
0
+
i
T
1
+
i
2
T
2
+
…
)
=
(
2016
C
0
x
2016
i
0
+
2016
C
1
x
2015
i
1
…
)
=
(
i
+
x
)
2016
(
T
0
−
i
T
1
+
i
2
T
2
−
i
3
T
3
…
)
=
(
2016
C
0
(
i
x
)
2016
+
2016
C
1
(
i
x
)
2015
…
)
=
(
1
+
i
x
)
2016
∴
(
T
0
+
i
T
1
+
i
2
T
2
+
⋯
+
i
2016
T
2016
)
(
T
0
−
i
T
1
+
i
2
T
2
−
i
3
T
3
…
)
=
(
i
+
x
)
2016
×
(
1
+
i
x
)
2016
=
[
(
i
+
x
)
(
1
+
i
x
)
]
2016
=
[
i
−
x
+
x
+
i
x
2
]
2016
=
(
1
+
x
2
)
2016
If
C
0
,
C
1
,
C
2
,
.
.
.
.
,
C
n
are binomial coefficients of order
n
, then the value of
C
1
2
+
C
3
4
+
C
5
6
+
.
.
.
.
=
Report Question
0%
2
n
+
1
n
+
1
0%
2
n
−
1
n
+
1
0%
2
n
+
1
n
−
1
0%
2
n
n
+
1
Explanation
(
1
+
x
)
n
=
C
0
+
C
1
x
+
C
2
x
2
+
C
3
x
3
+
.
.
.
.
.
.
C
n
x
n
Integrating, We have
(
1
+
x
)
n
+
1
n
+
1
=
C
0
x
+
C
1
x
2
2
+
C
2
x
3
3
+
.
.
.
.
.
C
n
x
n
+
1
n
+
1
Putting,
x
=
1
(
2
)
n
+
1
n
+
1
=
C
0
+
C
1
1
2
+
C
2
1
3
+
.
.
.
.
Putting,
x
=
−
1
(
0
)
n
+
1
n
+
1
=
−
C
0
+
C
1
1
2
+
C
2
(
−
1
)
3
+
.
.
.
.
Adding above two equations we have,
(
2
)
n
+
1
n
+
1
=
2
(
C
1
2
+
C
3
4
.
.
.
.
.
)
⟹
(
2
)
n
n
+
1
=
C
1
2
+
C
3
4
.
.
.
.
.
Hence, option D is correct.
The total number of terms in the expansion of
(
x
+
a
)
47
−
(
x
−
a
)
47
after simplification is
Report Question
0%
24
0%
47
0%
48
0%
96
Explanation
(
x
+
a
)
47
−
(
x
−
a
)
47
When we expand the above equation using binomial expansion
(
x
+
y
)
n
=
(
n
∑
k
=
0
n
C
k
x
k
y
n
−
k
)
So the above equation becomes
(
x
+
a
)
47
=
(
47
∑
k
=
0
47
C
k
x
k
a
47
−
k
)
(
x
−
a
)
47
=
(
47
∑
k
=
0
47
C
k
x
k
(
−
a
)
47
−
k
)
(
x
+
a
)
47
⇒
There are 48 terms in the expansion and all are positive
(
x
−
a
)
47
⇒
There are 48 terms in the expansion
The terms with odd powers of a will be cancelled and those with even powers of a will add up.
24 terms will be positive and 24 negative in the expansion of
(
x
−
a
)
47
48 terms positive-[24 terms negative and 24 terms positive]
=
48
t
e
r
m
s
p
o
s
i
t
i
v
e
+
24
t
e
r
m
s
n
e
g
a
t
i
v
e
+
24
t
e
r
m
s
p
o
s
i
t
i
v
e
=
24
terms
Let
(
(
1
+
x
)
+
x
2
)
9
=
a
0
+
a
1
x
+
a
2
x
2
+
.
.
.
.
.
+
a
18
x
18
. Then
Report Question
0%
a
0
+
a
2
+
.
.
.
.
.
+
a
18
=
a
1
+
a
3
+
.
.
.
.
.
+
a
17
0%
a
0
+
a
2
+
.
.
.
.
.
+
a
18
is even
0%
a
0
+
a
2
+
.
.
.
.
.
+
a
18
is divisible by
9
0%
a
0
+
a
2
+
.
.
.
.
.
+
a
18
is divisible by
3
but not by
9
Explanation
Given :
(
(
1
+
x
)
+
x
2
)
9
=
a
0
+
a
1
x
+
a
2
x
2
+
.
.
.
.
.
+
a
18
x
18
.....
(
i
)
Put
x
=
1
in
(
i
)
, we get
(
1
+
1
+
1
)
9
=
a
0
+
a
1
+
a
2
+
.
.
.
.
.
+
a
18
3
9
=
a
0
+
a
1
+
a
2
+
.
.
.
.
.
+
a
18
....
(
i
i
)
Put
x
=
−
1
in
(
i
)
, we get
(
1
−
1
+
1
)
9
=
a
0
−
a
1
+
a
2
−
a
3
+
a
4
−
.
.
.
.
.
−
a
17
+
a
18
1
=
a
0
−
a
1
+
a
2
−
a
3
+
a
4
−
.
.
.
.
.
−
a
17
+
a
18
.....
(
i
i
i
)
Adding
(
i
i
)
and
(
i
i
i
)
, we get
3
9
+
1
=
a
0
+
a
1
+
.
.
.
.
+
a
18
+
a
0
−
a
1
+
a
2
−
.
.
.
.
.
−
a
17
+
a
18
=
2
a
0
+
2
a
2
+
2
a
4
+
.
.
.
.
+
2
a
18
⟹
2
(
a
0
+
a
2
+
a
4
+
.
.
.
.
+
a
18
)
=
3
9
+
1
⟹
a
0
+
a
2
+
.
.
.
.
.
+
a
18
=
3
9
+
1
2
→
e
v
e
n
Hence,
a
0
+
a
2
+
a
4
+
.
.
.
.
+
a
18
is even.
Let
n
≥
5
and
b
≠
0
. In the binomial expansion of
(
a
−
b
)
n
, the sum of the 5th and 6th terms is zero then
a
/
b
equals
Report Question
0%
5
n
−
4
0%
1
5
(
n
−
4
)
0%
n
−
5
6
0%
n
−
4
5
Explanation
Solution:
Given that:
n
≥
5
,
b
≠
0
,
given expansion is
(
a
−
b
)
n
and
t
5
+
t
6
=
0
To find:
a
b
=
?
Solution:
t
5
=
n
C
4
a
4
(
−
b
)
(
n
−
4
)
and
t
6
=
n
C
5
a
5
(
−
b
)
(
n
−
5
)
∵
t
5
+
t
6
=
0
∴
n
C
4
a
4
(
−
b
)
(
n
−
4
)
+
n
C
5
a
5
(
−
b
)
(
n
−
5
)
=
0
⟹
n
C
5
a
=
n
C
4
b
⟹
a
b
=
n
C
4
5
C
5
=
n
!
4
!
(
n
−
4
)
!
n
!
5
!
(
n
−
5
)
!
=
5
n
−
4
Hence, A is the correct answer.
In the expansion of
(
3
x
−
1
x
2
)
10
, the
5
t
h
term from the end is
Report Question
0%
16486
x
8
0%
17010
x
8
0%
13486
x
8
0%
None of these
Explanation
There are
11
terms in the expansion of
(
3
x
−
1
x
2
)
10
Therefore,
5
t
h
term from the end.
=
(
10
−
5
+
2
)
t
h
term from beginning
=
T
7
=
10
C
6
(
3
x
)
10
−
6
(
−
1
x
2
)
6
=
10
!
6
!
4
!
3
4
x
4
1
x
12
=
17010
x
8
The coefficient of
x
49
in the product
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
.
.
.
.
(
x
−
50
)
is
Report Question
0%
−
2250
0%
−
1275
0%
1275
0%
2250
0%
−
49
Explanation
Coefficient of
x
49
in product of
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
.
.
.
.
.
.
.
.
(
x
−
50
)
We know that,
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
.
.
.
.
.
.
.
.
(
x
−
n
)
=
x
n
−
(
1
+
2
+
3
+
.
.
.
.
+
n
)
x
n
−
1
+
.
.
.
.
.
.
.
Coefficient of
x
49
=
−
(
1
+
2
+
3
+
.
.
.
+
50
)
=
−
50
×
51
2
=
−
1275
Hence, B is the correct option.
If
C
0
,
C
1
,
C
2
,
C
3
,
.
.
.
.
are binomial coefficients in the expansion of
(
1
+
x
)
n
, then
C
0
3
−
C
1
4
+
−
+
.
.
.
is equal to :
Report Question
0%
1
n
+
1
−
2
n
+
2
+
1
n
+
3
0%
1
n
+
1
+
2
n
+
2
−
3
n
+
3
0%
1
n
+
2
−
1
n
+
1
+
1
n
+
3
0%
2
n
+
1
−
1
n
+
2
+
2
n
+
3
0%
1
n
+
2
−
2
n
+
1
+
3
n
+
3
Explanation
We know
(
1
−
x
)
n
=
C
0
−
C
1
x
+
C
2
x
2
−
.
.
.
+
(
−
1
)
n
C
n
⋅
x
n
On multiplying both sides bby
x
2
, we get
(
1
−
x
)
n
x
2
=
C
0
x
2
−
C
1
x
3
+
C
2
x
4
−
.
.
.
+
On integrating both sides by taking limit
0
to
1
.
Therefore,
∫
1
0
(
1
−
x
)
n
x
2
d
x
=
∫
1
0
(
C
0
x
2
−
C
1
x
3
+
C
2
x
4
+
.
.
.
.
)
d
x
∫
1
0
x
n
(
1
−
x
)
2
d
x
=
[
C
0
x
3
3
−
C
1
x
4
4
+
C
2
x
5
5
−
.
.
.
]
1
0
⇒
∫
1
0
x
n
(
1
+
x
2
−
2
x
)
d
x
=
C
0
3
−
C
1
4
+
C
2
5
−
.
.
.
Here
C
0
3
−
C
1
4
+
C
2
5
−
.
.
.
=
[
x
n
+
1
n
+
1
+
x
n
+
3
n
+
3
−
2
x
n
+
2
n
+
2
]
1
0
=
[
1
n
+
1
+
1
n
+
3
−
2
n
+
2
]
The value of
r
for which the coefficients of
(
r
−
5
)
th and
(
3
r
+
1
)
th terms in the expansion of
(
1
+
x
)
1
/
2
are equal, is
Report Question
0%
4
0%
9
0%
12
0%
None of these
Explanation
Since, coefficient of
(
r
−
5
th term
=
coefficient of
=
(
3
r
+
1
)
th term
12
C
r
−
6
=
12
C
3
r
⇒
r
−
6
=
3
r
or
12
−
r
+
6
=
3
r
→
2
r
=
−
6
or
4
r
=
18
⇒
r
=
−
3
or
r
=
18
4
Hence no value of
r
exist, because
r
neither be negative nor in fraction
If
(
1
+
x
+
x
2
)
n
=
1
+
a
1
x
+
a
2
x
2
+
.
.
.
+
a
2
n
x
2
n
, then
2
a
1
−
3
a
2
+
.
.
.
−
(
2
n
+
1
)
a
2
n
is equal to
Report Question
0%
n
0%
−
n
0%
n
+
1
0%
−
n
−
1
0%
−
n
+
1
Explanation
Given,
(
1
+
x
+
x
2
)
n
=
1
+
a
1
x
+
a
2
x
2
+
.
.
.
+
a
2
n
x
2
n
⇒
x
(
1
+
x
+
x
2
)
n
=
x
+
a
1
x
2
+
a
2
x
3
+
.
.
.
+
a
2
n
x
2
n
+
1
On diffferentiating w.r.t.
x
, we get
(
1
+
x
+
x
2
)
n
+
x
⋅
n
(
1
+
x
+
x
2
)
n
−
1
(
1
+
2
x
)
=
1
+
2
a
1
x
+
3
a
2
x
2
+
.
.
.
+
a
2
n
⋅
(
2
n
+
1
)
x
2
n
On putting
x
=
−
1
, we get
(
1
−
1
+
1
)
n
−
n
(
1
−
1
+
1
)
n
−
1
(
1
−
2
)
=
1
−
2
a
1
+
3
a
2
+
.
.
.
+
a
2
n
(
2
n
+
1
)
⇒
1
−
n
(
−
1
)
=
1
−
2
a
1
+
3
a
2
+
.
.
.
+
a
2
n
(
2
n
+
1
)
⇒
2
a
1
−
3
a
2
.
.
.
.
=
(
2
n
+
1
)
a
2
n
=
−
n
The middle term in the expansion of
(
10
x
+
x
10
)
10
is
Report Question
0%
10
C
5
0%
10
C
6
0%
10
C
5
1
x
10
0%
10
C
5
x
10
0%
10
C
5
10
10
Explanation
MIddle term of the expansion of
(
10
x
+
x
10
)
10
is
T
6
.
T
6
=
10
C
5
(
10
x
)
5
(
x
10
)
5
T
6
=
10
C
5
Hence, A is the correct option.
If the term free from
x
in the expansion of
(
√
x
−
k
x
2
)
10
is
405
, then the value of
k
is
Report Question
0%
±
1
0%
±
3
0%
±
4
0%
±
2
Explanation
General term in the expansion of
(
√
x
−
k
x
2
)
T
r
+
1
=
10
C
r
(
√
x
)
10
−
r
(
−
k
x
2
)
r
=
10
C
r
x
10
−
r
2
⋅
(
−
k
)
r
⋅
x
−
2
r
=
10
C
r
(
−
k
)
r
x
(
10
−
5
r
2
)
The term is free from
x
.
Put
10
−
5
r
2
=
0
⇒
r
=
2
Now,
10
C
2
(
−
k
)
2
=
405
⇒
10
×
9
1
×
2
⋅
k
2
=
405
⇒
k
2
=
405
45
=
9
⇒
k
=
±
3
.
Sum of coefficients of the last
6
terms in the expansion of
(
1
+
x
)
11
when the expansion is in ascending powers of
x
, is
Report Question
0%
2048
0%
32
0%
512
0%
64
0%
1024
Explanation
(
1
+
x
)
′
′
=
11
∑
r
=
0
n
C
r
(
1
)
n
−
r
(
x
)
r
last
6
terms of this expansion are
′
′
C
11
+
′
′
C
10
+
′
′
C
q
+
′
′
C
8
+
′
′
C
7
+
′
′
C
6
using
n
C
r
+
n
C
r
+
1
=
n
+
1
C
r
+
1
=
12
C
11
+
12
C
9
+
12
C
7
=
12
+
220
+
792
=
1024
If
C
0
,
C
1
,
C
2
,
…
,
C
15
are binomial coefficients in
(
1
+
x
)
15
, then
C
1
C
0
+
2
C
2
C
1
+
3
C
3
C
2
+
⋯
+
15
C
15
C
14
is equal to
Report Question
0%
60
0%
120
0%
64
0%
124
0%
144
Explanation
We know that,
n
C
r
n
C
r
−
1
=
n
−
(
r
−
1
)
r
⇒
r
⋅
n
C
r
n
C
r
−
1
=
n
+
1
−
r
⇒
n
∑
r
=
1
r
⋅
16
C
r
16
C
r
−
1
=
16
∑
r
=
1
(
16
−
r
)
16
×
15
−
n
∑
r
=
1
r
=
16
×
15
−
15
×
16
2
=
240
−
120
=
120
The middle term in the expansion of
(
1
+
x
)
2
n
is
Report Question
0%
1.3.5....
(
2
n
−
1
)
2
n
n
!
0%
1.2.3....
(
2
n
−
1
)
2
n
x
n
n
!
0%
1.3.5....
(
2
n
−
1
)
x
n
n
!
0%
1.3.5....
(
2
n
−
1
)
2
n
x
n
n
!
Explanation
Given expansion is
(
1
+
x
)
2
n
,
The total number of terms are
2
n
+
1
which is an odd number,
∴
there is only one middle term which is
(
2
n
n
)
,
(
2
n
n
)
=
(
2
n
)
!
n
!
n
!
⟹
[
1.3.5...
(
2
n
−
1
)
]
[
2
n
n
!
]
n
!
n
!
x
n
=
[
1.3.5...
(
2
n
−
1
)
]
[
2
n
]
n
!
x
n
If
n
ϵ
N
and
(
1
+
4
x
+
4
x
2
)
n
=
r
=
2
n
∑
r
=
0
a
r
x
r
then value of
2
n
∑
r
=
0
a
2
r
equals
Report Question
0%
9
n
−
1
0%
9
n
+
1
0%
3
n
+
1
0%
3
n
−
1
Explanation
∵
1
+
4
x
+
4
x
2
=
(
1
+
2
x
)
2
∴
(
1
+
2
x
)
2
n
=
2
n
∑
r
=
0
a
r
x
r
putting
x
=
1
3
2
n
=
2
n
∑
r
=
0
a
r
=
a
0
+
a
1
+
a
2
+
a
3
+
.
.
.
.
.
+
a
2
n
.
.
.
.
(
i
)
and putting
x
=
−
1
1
=
r
=
2
n
∑
r
=
0
a
r
(
−
1
)
r
=
a
0
−
a
1
+
a
2
−
a
3
+
.
.
.
.
+
a
2
n
.
.
.
(
i
i
)
Now adding (i) and (ii) we get
2
(
a
0
+
a
2
+
a
4
+
.
.
.
.
+
a
2
n
)
=
3
2
n
+
1
⇒
2
r
=
n
∑
r
=
0
a
2
r
=
9
n
+
1
Hence choice (b) is correct answer.
The sum of the co-efficients of all odd degree terms in the expansion of
(
x
+
√
x
3
−
1
)
5
+
(
x
−
√
x
3
−
1
)
5
,
(
x
>
1
)
is
Report Question
0%
1
0%
2
0%
−
1
0%
0
Explanation
(
x
+
√
x
3
−
1
)
5
+
(
x
−
√
x
3
−
1
)
5
(
a
+
b
)
5
=
a
5
+
5
a
4
b
+
10
a
3
b
2
+
10
a
2
b
3
+
5
a
b
4
+
b
5
−
−
−
−
−
−
−
(
i
)
(
a
−
b
)
5
=
a
5
−
5
a
4
b
+
10
a
3
b
2
−
10
a
2
b
3
+
5
a
b
4
−
b
5
−
−
−
−
−
−
−
(
i
i
)
(
a
+
b
)
5
+
(
a
−
b
)
5
=
2
[
a
5
+
10
a
3
b
2
+
5
a
b
4
]
=
2
[
x
5
+
10
x
3
(
x
3
−
1
)
+
5
x
(
x
3
−
1
)
4
]
=
2
[
x
5
+
10
x
6
−
10
x
3
+
5
x
(
x
6
−
2
x
3
+
1
)
]
=
2
x
5
+
20
x
6
−
20
x
3
+
10
x
7
−
20
x
4
+
10
x
Here all the co-efficient of the above equation are odd terms
coefficients of even terms are cancelled out.
So, sum of coefficients
⇒
2
+
20
−
20
+
10
−
20
+
10
=
2
If
C
r
denotes the binomial coefficient
n
C
r
then
(
−
1
)
C
2
0
+
2
C
2
1
+
5
C
2
2
+
.
.
.
.
.
.
(
3
n
−
1
)
C
2
n
=
Report Question
0%
(
3
n
−
2
)
2
n
C
n
0%
(
3
n
−
2
2
)
2
n
C
n
0%
(
5
+
3
n
)
2
n
C
n
0%
(
3
n
−
5
2
)
2
n
C
n
+
1
Explanation
The general term of above series is
T_{r} = (3r-1)C_{r}^{2}
(-1)C_{0}^{2}+2C_{1}^{2}+5C_{2}^{2}+...+(3n-1)C_{n}^{2}
=\sum_{r=0}^{r=n}(3r-1)C_{r}^{2}
=3\sum_{r=0}^{r=n}rC_{r}^{2} - \sum_{r=0}^{r=n}C_{r}^{2}
=3S_{1} - S_{2}
The binomial expansion of
(1+x)^n
and
(1+\dfrac{1}{x})^n
is given by
(1+x)^n = C_{0}+C_{1}x+C_{2}x^{2}+....+C_{n}x^{n} = \sum_{r=0}^{r=n}C_{r}x^{r}
....[1]
(1+\dfrac{1}{x})^n = C_{0}+C_{1}\dfrac{1}{x}+C_{2}\dfrac{1}{x^{2}}+....+C_{n}\dfrac{1}{x^{n}} = \sum_{r=0}^{r=n}C_{r}\dfrac{1}{x^{r}}
....[2]
To find the value of
S_{2}
, we can multiply above two equations and then compare the coefficients of terms independent of x.
Multiplying equation [1] and [2], we get
(1+x)^n (1+\dfrac{1}{x})^{n} = \sum_{r=0}^{r=n}C_{r}x^{r}.\sum_{r=0}^{r=n}C_{r}\dfrac{1}{x^{r}}
\implies \dfrac{(1+x)^{2n}}{x^{n}} = \sum_{r=0}^{r=n}C_{r}x^{r}.\sum_{r=0}^{r=n}C_{r}\dfrac{1}{x^{r}}
\implies \dfrac{(1+x)^{2n}}{x^{n}}=C_{0}^{2}+C_{1}^{2}+C_{2}^{2}+...+C_{n}^{2}
+ terms containing x
Therefore, comparing coefficients of
x^{0}
in L.H.S and R.H.S, we get
Coefficients of
x^{0}
in R.H.S = C
oefficients of
x^{0}
in L.H.S
\implies C_{0}^{2}+C_{1}^{2}+C_{2}^{2}+...+C_{n}^{2}=
coefficient of
x^{0}
in
\dfrac{(1+x)^{2n}}{x^{n}}
\implies C_{0}^{2}+C_{1}^{2}+C_{2}^{2}+...+C_{n}^{2}= ^{2n}C_{n} = S_{2}
....[3]
(1+\dfrac{1}{x})^n = C_{0}+C_{1}\dfrac{1}{x}+C_{2}\dfrac{1}{x^{2}}+....+C_{n}\dfrac{1}{x^{n}}
Differentiating with respect to x, we get
\implies n(1+\dfrac{1}{x})^{n-1}(\dfrac{-1}{{x^{2}}}) = 0+C_{1}\dfrac{-1}{x^{2}}+C_{2}\dfrac{-2}{x^{3}}+....+C_{n}\dfrac{-n}{x^{n+1}}
\implies n(1+\dfrac{1}{x})^{n-1}(\dfrac{1}{{x}}) = C_{1}\dfrac{1}{x}+C_{2}\dfrac{2}{x^{2}}+....+C_{n}\dfrac{n}{x^{n}}
....[4]
To find the value of
S_{1}
, we can multiply equations [1] and [4], and then compare the coefficients of terms independent of x.
Multiplying equation [1] and [4], we get
n(1+\dfrac{1}{x})^{n-1}(\dfrac{1}{{x}})(1+x)^n=(C_{1}\dfrac{1}{x}+C_{2}\dfrac{2}{x^{2}}+....+C_{n}\dfrac{n}{x^{n}})( C_{0}+C_{1}x+C_{2}x^{2}+....+C_{n}x^{n} )
\implies n\dfrac{(x+1)^{2n-1}}{x^{n}}=(C_{1}\dfrac{1}{x}+C_{2}\dfrac{2}{x^{2}}+....+C_{n}\dfrac{n}{x^{n}})( C_{0}+C_{1}x+C_{2}x^{2}+....+C_{n}x^{n} )
Therefore, comparing coefficients of
x^{0}
in L.H.S and R.H.S, we get
Coefficients of
x^{0}
in R.H.S = C
oefficients of
x^{0}
in L.H.S
\implies C_{1}^{2}+2C_{2}^{2}+3C_{3}^{2}+...+nC_{n}^{2}=
coefficient of
x^{0}
in
n\dfrac{(x+1)^{2n-1}}{x^{n}}
\implies C_{1}^{2}+2C_{2}^{2}+3C_{3}^{2}+...+nC_{n}^{2}= n(^{2n-1}C_{n}) = S_{1}
Therefore,
3S_{1} - S_{2} = 3n(^{2n-1}C_{n})-^{2n}C_{n}
= 3n\dfrac{(^{2n}C_{n})}{2}-^{2n}C_{n}
= (\dfrac{3n-2}{2})^{2n}C_{n}
Hence, the answer is option (B)
Given
(1-2x+5x^2-10x^3)(1+x)^n=1+a_1x+a_2x^2+...
and that
a_1^2=2a_2
then the value of
n
is-
Report Question
0%
6
0%
2
0%
5
0%
3
Explanation
Given that
\left( 1-2x+5{ x }^{ 2 }-10{ x }^{ 3 } \right) { \left( 1+x \right) }^{ n }=1+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+...
Given that
{ { a }_{ 1 } }^{ 2 }=2{ a }_{ 2 }
Here
{ a }_{ 1 }
is the coefficient of
x
and
{ a }_{ 2 }
is the coefficient of
{ x }^{ 2 }
,
\Longrightarrow { a }_{ 1 }=n-2
and
\Longrightarrow { a }_{ 2 }=-2n+{ C }_{ 2 }+5
Substituting these values in the given relation we get,
\Longrightarrow { \left( n-2 \right) }^{ 2 }=2\left( 5-2n+\frac { n\left( n-1 \right) }{ 2 } \right) \\ \Longrightarrow n=6\\
If
{ C }_{ 0 },{ C }_{ 1 },{ C }_{ 2 },.....{ C }_{ r }
are binomial coefficients in the expansion of
{(1+x)}^{n}
then
{ C }_{ 1 }-\cfrac { { C }_{ 2 } }{ 2 } +\cfrac { { C }_{ 3 } }{ 3 } -\cfrac { { C }_{ 4 } }{ 4 } +....{ \left( -1 \right) }^{ n-1 }\cfrac { { C }_{ n } }{ n }
equals
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0%
\sum _{ r=1 }^{ n }{ { \left( -1 \right) }^{ n-1 } } \cfrac { { C }_{ r } }{ r }
0%
\sum _{ r=1 }^{ n }{ \cfrac { 1 }{ r } }
0%
\sum _{ r=2 }^{ n }{ \cfrac { 1 }{ r-1 } }
0%
None of these
Explanation
Consider
{ (1-x) }^{ n }={ C }_{ 0 }-{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }-{ C }_{ 3 }{ x }^{ 3 }+......
\Rightarrow 1-{ (1-x) }^{ n }={ C }_{ 1 }x-{ C }_{ 2 }{ x }^{ 2 }+{ C }_{ 3 }{ x }^{ 3 }-......
integrating both side w.r.t
x
with limit
0
to
1
\displaystyle \therefore \int _{ 0 }^{ 1 }{ \left( { C }_{ 1 }x-{ C }_{ 2 }{ x }^{ 2 }+{ C }_{ 3 }{ x }^{ 3 }-..... \right) dx } =\int _{ 0 }^{ 1 }{ \cfrac { 1-{ (1-x) }^{ n } }{ 1-(1-x) } } dx
\displaystyle \Rightarrow \cfrac { { C }_{ 1 } }{ 1 } -\cfrac { { C }_{ 2 } }{ 2 } +\cfrac { { C }_{ 3 } }{ 3 } +....{ (-1) }^{ n }-1\cfrac { { C }_{ n } }{ n } =\int _{ 0 }^{ 1 }{ \cfrac { 1-{ x }^{ n } }{ 1-x } } dx\quad
\displaystyle \left( Using\quad \int _{ 0 }^{ a }{ f(x) } dx=\int _{ 0 }^{ a }{ f(a- } x)dx \right)
\displaystyle =\int _{ 0 }^{ 1 }{ \left( 1+x+{ x }^{ 2 }+...+{ x }^{ n-1 } \right) dx }
=1+\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 4 } +...\cfrac { 1 }{ n } =\sum _{ r=1 }^{ n }{ \cfrac { 1 }{ r } }
Prove that the coefficient of middle term in the expansion of
( 1 + x ) ^{2n}
is equal to the sum of the coefficient of two middle terms in
(1 + x)^{2n - 1}
.
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0%
True
0%
False
Explanation
Statement is true, so option
A
is correct.
Proof :- Middle term of
(1 + X)^{2n}
Since
2n
is even middle term =
\left(\dfrac{2n}{2} + 1\right)^{th} = (n + 1)^{th}
term =
T_{n + 1}
we know that general & term of expansion
(a + b)^n
is -
T_{n + 1} = \,^nC_r \, a^{n - r} b^r
here,
r = n , n = 2n , a = 1 , b = x
putting values
T_{nH} = \, ^{2n}C_n (1)^{2n - n} (x)^n
T_{nH} = \,^{2n}C_n (x)^n......(A)
Here, coefficient is
^{2n} C_n
\Rightarrow
middle term of
(1 + x)^{2n - 1}
Since,
(2n - 1)
is add
There will be 2 middle terms
\dfrac{(2n - 1) + 1}{2}
and
\left(\dfrac{(2n - 1) + 1}{2} + 1 \right)
term
= n^{th}
term and
(n + 1)^{th}
term
= T_n
and
T_{n + 1}
we know that general term for expansion
(a + b)^n
T_{rH} = \,^nC_r \, a^{n - r} \, b^r
for
T_n
in
(1 + x)^{2n - 1}
Putting
r = n - 1, n = 2n - 1, a = 1 \, \& \, b = x
T_n = \, ^{2n - 1} C_{n - 1} \, (1)^{2n - 1 - ( n - 1)} x_{( n- 1)} = \, ^{2n - 1}C_{n - 1} x^{(n - 1)}
Coefficient pf middle term
(n^{th} \, term) = \, ^{2n - 1}C_{n - 1}
T_{nH} = \, ^{2n - 1}C_n \, (1)^{2n - 1 - n} x^{(n)} = \, ^{2n - 1} C_{n} \, (X)^n
Sum of the coefficent of the middle term in the expansion of
(1 + X)^{2n - 1}
= \,^{2n -1}C_{n - 1} + \, ^{2n - 1}C_n
= \dfrac{(2n- 1)!}{(2n - 1 - n + 1)! (n -1)!} + \dfrac{(2n - 1)!}{(2n - 1 - n)! n!}
= \dfrac{(2n - 1)!}{n! (n -1)!} + \dfrac{(2n - 1)!}{(n - 1)! n!}
= \dfrac{2 (2n - 1)!}{n! (n - 1)!} \times \dfrac{n}{n}
= \dfrac{2n (2n - 1)!}{n! n (n - 1)!} = \dfrac{(2n)!}{n! \, n!} = \, ^{2n}C_n .....(B)
Equation
(A)
= Equation
(B)
State true or false.
The middle term in the expansion of
\left(\dfrac{x}{a} \, - \, \dfrac{a}{x}\right)^{10}=-252
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0%
True
0%
False
Find the term of the expansion of
\displaystyle\, \left ( \sqrt[3]{x^{-2}} + x \right )^7
containing
x
in the second power.
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0%
T_4
0%
T_5
0%
T_6
0%
T_7
Explanation
Note:- The general term or
(r+1)^{th}
term of
(a+b)^{n}
is
{T}_{r+1}=^{ n }{ C }_{ r }{a}^{n-r}{b}^{r}
{T}_{r+1}=^{ 7 }{ C }_{ r }\left( \sqrt [ 3 ]{ { x }^{ -2 } } \right) ^{ 7-r }\left(x \right)^{r}
\Rightarrow {T}_{r+1}= ^{ 7 }{ C }_{ r }\left({x}^{-2}\right)^{{7-r}/{3}}{x}^{r}
\Rightarrow {T}_{r+1}=^{ 7 }{ C }_{ r } {x}^{{-2}/{3}(7-r)+r}
Now, if power of
x
must be
2
, then
r
is:-
\therefore {x}^{{-2}/{3}(7-r)+r}= {x}^{2}
\Rightarrow \cfrac{-2}{3}(7-r)+r=2
\Rightarrow \cfrac{-14}{3}+\cfrac{2}{3}r+r=2
\Rightarrow \cfrac{5r}{3}= \cfrac{20}{3}
\Rightarrow r=4
Therefore
{T}_{4+1}={T}_{5}
is the term which contains
{x}^{2}
Sum of coefficients in the expression of
(x+2y+z)^{10}
is
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0%
2^{10}
0%
3^{10}
0%
1
0%
None of these
Explanation
Sum of = coeff.
4^{10}
The
6^{th}
coefficient in the expansion of
\left (2x^2 - \dfrac {1}{3x^2}\right)^{10}
Report Question
0%
-\dfrac{986}{27}
0%
\dfrac{986}{27}
0%
\dfrac{896}{27}
0%
- \dfrac{896}{27}
Explanation
{ \left( 2{ x }^{ 2 }-\dfrac { 1 }{ { { 3x }^{ 2 } } } \right) }^{ 10 }
6^{th}
coefficient
\rightarrow
5^{th}
term
={ ^{ 10 }{ C } }_{ 5 }{ \left( 2{ x }^{ 2 } \right) }^{ 5 }{ \left( \dfrac { -1 }{ { { 3x }^{ 2 } } } \right) }^{ 5 }
={ ^{ 10 }{ C } }_{ 5 }\times 2^5\times \dfrac{-1}{3^5}
=-\dfrac{252\times 32}{243}
=-\dfrac{896}{27}.
Hence, the answer is
-\dfrac{896}{27}.
Prove that
C_0+C_1+C_2+.....C_n=2^n
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0%
2^n
0%
2^{n-1}
0%
2^{n+1}
0%
None of the above.
Explanation
(1+x)^n=^nC_0+^nC_1x+\cdots+^nC_nx^n
Let
x=1
\Rightarrow C_0+C_1.1+C_2.1^2+\cdots+C_n.1^n=(1+1)^n=2^n
\Rightarrow C_0+C_1+C_2.1+\cdots+C_n=(1+1)^n=2^n
In the expansion of
(\sqrt[5]{3}+\sqrt[7]{2})^{24}
, the rational term is
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T_{14}
0%
T_{16}
0%
T_{15}
0%
T_7
the coefficients of
x^{49}
in the polynomial.
\left (x \, - \, \dfrac{C_1}{C_0}\right) \, \left (x \, - \, 2^2 \, \dfrac{C_2}{C_1}\right) \, \left (x \, - \, 3^2 \, \dfrac{C_3}{C_2}\right) \, ..... \, \, \left (x \, - \, 50^2 \, \dfrac{C_{50}}{C_{49}}\right)
is
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0%
50\displaystyle \sum _{ r=1 }^{ 50 }{ r-{ r }^{ 2 } }
0%
51\displaystyle \sum _{ r=1 }^{ 50 }{ { r }^{ 2 } } -\displaystyle \sum _{ r=1 }^{ 50 }{ r }
0%
51\displaystyle \sum _{ r=1 }^{ 50 }{ { r } } -\displaystyle \sum _{ r=1 }^{ 50 }{ r^{ 2 } }
0%
none\ of\ these
Explanation
We Know that
(x \, - \, a) \, (x \, - \, b) \,(x \, - \, c).....50
factors
= \, x^{50}\, x^{49} \, . S_1 \, + x^{48} \, S_2 \, -....
The coefficient of
x^{49}
is
S_1 \, = \,- \,\left [ \dfrac{C_1}{C_0} \, + \, 2^2 \, \dfrac{C_2}{C_1} \,+ \,3^2 \, \dfrac{C_3}{C_2}\right] \, + .....r^2 \,\dfrac{C_r}{C_{r \, - \, 1}}]
Now \, r^2 \, .\dfrac{C_r}{C_{r \, - \, 1}} \, = \, r^2 \, .\dfrac{n \, -r \, + \, 1}{r}
= r \, (51 \, - \, r) \, = \, 51r \, - r^2
\, \therefore \, \sum_{r \, = \, 1}^{50} \, r^2 \, \dfrac{c_r}{C_{r \, - \, 1}} \, 51 \, \sum_{r \, = \, 1}^{50} \, r \, - \, \sum_{r \, = \, 1}^{50} \, r^2
= \, 51. \, \dfrac{N (N \, + \, 1)}{2}\, - \, \dfrac{N(N \, + \, 1)\, (2N \, + \, 1)}{6}
= \, 51. \, \dfrac{50. \, 51)}{2}\, - \, \dfrac{50. \ 51. \, (101)}{6}
= \,\dfrac{1}{6} \, 50. \, 51 \, \left [ 3 \, \times \, 51 \, - \, 101 \right ] \, = \, 25 \, \times \, 17 \, \times \, 52
If the sum of the co-efficient in the expansion of
(a+b)^n
is
1024
, then the greatest co-efficient in the expansion is
Report Question
0%
252
0%
352
0%
452
0%
552
Explanation
Sum of coefficient in the expansion of
(a+b)^{n}
is
1024
put
a=b=1,2^{n}=1024\implies n=10
The greatest coefficient is
^{10}C_{5}=252
If
n\ge 2
then
3.{ C }_{ 1 }-4.{ C }_{ 2 }+5.{ C }_{ 3 }-......+{ \left( -1 \right) }^{ n-1 }\left( n+2 \right) .{ C }_{ n }
is equal to
Report Question
0%
-1
0%
2
0%
-2
0%
1
Explanation
(1+x)^n=^nC_0+^nC_1x+^nC_2x^2+....+^nC_n x^n
x^2(1+x)^n=^nC_0 x^2+^nC_1x^3+^nC_2x^4+....+^nC_n x^{n+2}
differentiate w.r.t
x
2x(1+x)^n+nx^2(1+x)^{n-1}=2^nC_0 x+3 ^nC_1x^2+4 ^nC_2x^3+....+(n+2)^nC_n x^{n+1}
put
x=-1
0=-2+3 ^nC_1-4 ^nC_2+5 ^nC_3+.....+(-1)^{n-1}(n+2) ^nC_n
so
3 ^nC_1- 4 ^nC_2+5 ^nC_3+.....+(-1)^{n-1}(n+2) ^nC_n=2
The sum
^{10}C_3 + ^{11}C_3 + ^{12}C_3 + .... + ^{20}C_3
is equal to
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0%
^{21}C_4
0%
^{21}C_4 - ^{10}C_4
0%
^{21}C_4 - ^{11}C_4
0%
^{21}C_{17}
Explanation
^{10}C_{3} +\ ^{11}C_{3} +\ ^{12}C_{3} + --- +\ ^{20}C_{3}
....(i)
addition and subtraction
^{10}C_{4}
in equation (i)
^{10}C_{4} -\ ^{10}C_{4} +\ ^{10}C_{3} +\ ^{11}C_{3} + --- +\ ^{20}C_{3}
[we know that -
^nc_{r-1} +\ ^nc_{r} =\ ^{n+1}c_{r}]
^{10}C_{3} +\ ^{10}C_{4} +\ ^{11}C_{3} + --- +\ ^{20}C_{3} -\ ^{10}C_{4}
=\ ^{11}C_{4} +\ ^{11}C_{3} +\ ^{12}C_{3} + ---+\ ^{20}C_{3} -\ ^{10}C_{4}
=\ ^{12}C_{4} +\ ^{12}C_{3} + --- +\ ^{20}C_{3} -\ ^{10}C_{4}
=\ ^{13}C_{4} +\ ^{14}C_{3} + --- +\ ^{20}C_{3} -\ ^{10}C_{4}
Similarly
,
Continue
So, we get
^{21}C_{4} - \ ^{10}C_{4}
So, option (B) is correct
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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