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CBSE Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 6 - MCQExams.com
CBSE
Class 11 Engineering Maths
Principle Of Mathematical Induction
Quiz 6
If
n
is an odd positive integer, then
a
n
+
b
n
is divisible by
Report Question
0%
a
−
b
0%
a
+
b
0%
a
2
+
b
2
0%
none of these
Explanation
Let,
P
(
n
)
=
a
n
+
b
n
P
(
1
)
=
a
+
b
, which is divisible by
a
+
b
Now let
P
(
k
)
=
a
k
+
b
k
is divisible by
a
+
b
, where
k
is an odd integer.
⇒
a
k
+
b
k
=
(
a
+
b
)
f
(
a
,
b
)
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
(
1
)
Now,
P
(
k
+
2
)
=
a
k
+
2
+
b
k
+
2
=
a
2
[
(
a
+
b
)
f
(
a
,
b
)
−
b
k
]
+
b
k
+
2
=
a
2
f
(
a
,
b
)
(
a
+
b
)
−
a
2
b
k
+
b
k
+
2
from
(
1
)
=
a
2
f
(
a
,
b
)
(
a
+
b
)
−
b
k
(
a
2
−
b
2
)
=
(
a
+
b
)
[
a
2
f
(
a
,
b
)
−
b
k
(
a
−
b
)
]
, which is divisible by
(
a
+
b
)
Hence
a
n
+
b
n
is divisible by
(
a
+
b
)
for all odd positive integral
n
The value of
1
2
1.3
+
2
2
3
.
5
+
⋯
+
n
2
(
2
n
−
1
)
(
2
n
+
1
)
is
Report Question
0%
n
(
n
+
1
)
2
(
2
n
+
1
)
0%
n
(
n
−
1
)
2
(
2
n
−
1
)
0%
n
2
(
n
−
1
)
2
2
(
2
n
+
1
)
0%
none of these
Explanation
n
2
(
2
n
−
1
)
(
2
n
+
1
)
=
4
n
2
4
(
2
n
−
1
)
(
2
n
+
1
)
=
1
4
[
4
n
2
−
1
+
1
4
n
2
−
1
]
=
1
4
[
1
+
1
4
n
2
−
1
]
=
1
4
[
1
+
1
2
(
2
n
+
1
−
(
2
n
−
1
)
4
n
2
−
1
)
]
=
1
4
[
1
+
1
2
(
1
2
n
−
1
−
1
2
n
+
1
)
]
Consider
1
2
n
−
1
−
1
2
n
+
1
=
1
−
1
3
+
1
3
−
1
5
+
1
5
.
.
.
1
2
n
−
1
−
1
2
n
+
1
=
1
−
1
2
n
+
1
=
2
n
2
n
+
1
Hence applying summation gives us
n
4
+
1
8
∑
(
1
2
n
−
1
−
1
2
n
+
1
)
=
n
4
+
2
n
8
(
2
n
+
1
)
=
n
4
+
n
4
(
2
n
+
1
)
=
n
(
2
n
+
1
)
+
n
4
(
2
n
+
1
)
=
n
(
2
n
+
2
)
4
(
2
n
+
1
)
=
n
(
n
+
1
)
2
(
2
n
+
1
)
The value of
1
1.2.3
+
1
2.3.4
+
.
.
.
.
+
1
n
(
n
+
1
)
(
n
+
2
)
is
Report Question
0%
n
(
n
+
3
)
4
(
n
+
1
)
(
n
+
2
)
0%
n
(
n
+
1
)
(
n
+
2
)
0%
n
(
n
+
2
)
(
n
+
1
)
(
n
+
3
)
0%
All of these
Explanation
Let
P
(
n
)
:
1
a
.2
.
3
+
1
2.3.4
+
.
.
.
.
.
+
1
n
(
n
+
1
)
(
n
+
2
)
=
n
(
n
+
3
)
4
(
n
+
1
)
(
n
+
2
)
...... (1)
Step I: For
n
=
1
,
L.H.S. of
(
1
)
=
1
1.2.3
=
1
6
and R.H.S of
(
1
)
=
1.
(
1
+
3
)
4
(
1
+
1
)
(
1
+
2
)
=
1
6
Therefore P(1) is true.
Step II : Assume that P(k) is true, then
P
(
k
)
:
1
1.2.3
+
1
2.3.4
+
.
.
.
.
+
1
k
(
k
+
1
)
(
k
+
2
)
=
k
(
k
+
3
)
4
(
k
+
1
)
(
k
+
2
)
Step III : For
n
=
k
+
1
,
P
(
k
+
1
)
:
1
1.2.3
+
1
2.3.4
+
.
.
.
.
+
1
k
(
k
+
1
)
(
k
+
2
)
+
1
(
k
+
1
)
(
k
+
2
)
(
k
+
3
)
=
(
k
+
1
)
(
k
+
4
)
4
(
k
+
2
)
(
k
+
3
)
∴
L.H.S
=
1
1.2.3
+
1
2.3.4
+
.
.
.
.
.
+
1
k
(
k
+
1
)
(
k
+
2
)
+
1
(
k
+
1
)
(
k
+
2
)
(
k
+
3
)
=
k
(
k
+
3
)
4
(
k
+
1
)
(
k
+
2
)
+
1
(
k
+
1
)
(
k
+
2
)
(
k
+
3
)
(By assumption step)
=
k
(
k
+
3
)
2
+
4
4
(
k
+
1
)
(
k
+
2
)
(
k
+
3
)
=
k
3
+
6
k
2
+
9
k
+
4
4
(
k
+
1
)
(
k
+
2
)
(
k
+
3
)
=
(
k
+
1
)
2
(
k
+
4
)
4
(
k
+
1
)
(
k
+
2
)
(
k
+
3
)
=
(
k
+
1
)
(
k
+
4
)
4
(
k
+
2
)
(
k
+
3
)
=
R
.
H
.
S
Hence P(k + 1) is true. Hence by the principle of mathematical induction P(n) is true for all n
∈
N.
The value of
tan
−
1
(
1
3
)
+
tan
−
1
(
1
7
)
+
⋯
+
tan
−
1
(
1
n
2
+
n
+
1
)
is
Report Question
0%
tan
−
1
(
n
n
+
2
)
0%
tan
−
1
(
n
+
1
n
−
1
)
0%
tan
−
1
(
n
−
1
n
+
2
)
0%
tan
−
1
(
n
+
2
n
−
2
)
Explanation
T
n
=
tan
−
1
(
1
1
+
n
(
1
+
n
)
)
=
tan
−
1
(
(
n
+
1
)
−
n
1
+
n
(
1
+
n
)
)
=
tan
−
1
(
n
+
1
)
−
tan
−
1
n
Hence required sum is
=
∑
T
n
=
[
tan
−
1
(
2
)
−
tan
−
1
1
]
+
[
tan
−
1
(
3
)
−
tan
−
1
2
]
+
.
.
.
.
.
.
.
.
+
tan
−
1
(
n
)
−
tan
−
1
(
n
−
1
)
]
+
[
tan
−
1
(
n
+
1
)
−
tan
−
1
n
]
=
tan
−
1
(
n
+
1
)
−
tan
−
1
1
=
tan
−
1
(
n
+
1
−
1
1
+
(
n
+
1
)
.1
)
=
tan
−
1
(
n
n
+
2
)
By mathematical induction
p
n
+
1
+
(
p
+
1
)
2
n
−
1
is divisible by
Report Question
0%
p
2
+
p
+
1
0%
p
2
+
1
0%
p
+
1
0%
None of these
Explanation
Let
f
(
n
)
=
p
n
+
1
+
(
p
+
1
)
2
n
−
1
we have
f
(
1
)
=
p
2
+
p
+
1
which is divisible by
p
2
+
p
+
1
Now, assume that
f
(
m
)
is divisible by
p
2
+
p
+
1
∴
p
m
+
1
+
(
p
+
1
)
2
m
−
1
=
k
(
p
2
+
p
+
1
)
...(1)
f
(
m
+
1
)
=
p
m
+
2
+
(
p
+
1
)
2
m
+
2
−
1
=
p
m
+
2
+
(
p
+
1
)
2
m
−
1
.
(
p
+
1
)
2
=
p
m
+
2
+
[
k
(
p
2
+
p
+
1
)
−
p
m
+
1
]
(
p
+
1
)
2
=
p
m
+
2
−
(
p
+
1
)
2
p
m
+
1
+
k
(
p
+
1
)
2
(
p
2
+
p
+
1
)
=
p
m
+
1
(
p
−
p
2
−
2
p
−
1
)
+
k
(
p
+
1
)
2
(
p
2
+
p
+
1
)
=
−
(
p
2
+
p
+
1
)
[
−
k
(
p
+
1
)
2
+
p
m
+
1
]
Hence,
f
(
m
+
1
)
is divisible by
p
2
+
p
+
1
Hence by mathematical induction
f
(
n
)
is divisible by
p
2
+
p
+
1
for all
n
∈
N
If
a
1
=
1
,
a
n
+
1
=
1
n
+
1
a
n
,
∀
n
≥
1
, then
a
n
=
Report Question
0%
1
n
!
0%
1
(
n
+
2
)
!
0%
1
(
n
+
1
)
!
0%
none of these
Explanation
a
1
=
1
,
a
2
=
1
1
+
1
a
1
=
1
2
(
1
)
=
1
2
!
,
a
3
=
1
3
.
1
2
!
=
1
3
!
Similarly
a
n
=
1
n
.
1
(
n
−
1
)
!
=
1
n
!
3
+
13
+
29
+
51
+
79
+
.
.
.
to
n
terms
=
Report Question
0%
2
n
2
+
7
n
3
0%
n
2
+
5
n
3
0%
n
3
+
2
n
2
0%
none of these
Explanation
Let
S
n
=
3
+
13
+
29
+
51
+
79
+
⋯
+
n
t
e
r
m
s
∵
S
1
=
3
and
S
2
=
3
+
13
=
16
Similarly
S
3
=
3
+
13
+
29
=
45
By options, from
(
c
)
as we have
S
n
=
n
3
+
2
n
2
check by putting
n
=
1
,
S
1
=
1
3
+
2.1
2
=
3
S
2
=
2
3
+
2.2
2
=
16
and
S
3
=
3
3
+
2.3
2
=
45
Thus option
(
c
)
is correct
Using the principle of mathematical induction, find
t
a
n
α
+
2
t
a
n
2
α
+
2
2
t
a
n
2
2
α
+
.
.
.
.
to
n
terms:
Report Question
0%
t
a
n
α
−
2
n
⋅
t
a
n
(
2
n
⋅
α
)
0%
c
o
t
α
−
2
n
⋅
c
o
t
(
2
n
⋅
α
)
0%
s
e
c
α
−
2
n
⋅
s
e
c
(
2
n
⋅
α
)
0%
None of these
Explanation
Let P(n):
t
a
n
α
+
2
t
a
n
2
α
+
2
2
t
a
n
2
2
α
+
.
.
.
.
+
2
n
−
1
t
a
n
(
2
n
−
1
α
)
=
c
o
t
α
−
2
n
⋅
c
o
t
(
2
n
⋅
α
)
..... (1)
Step I: For
n
=
1
,
L.H.S. of
(
1
)
=
t
a
n
α
=
c
o
t
α
−
c
o
t
α
+
t
a
n
α
=
c
o
t
α
−
(
c
o
t
α
−
t
a
n
α
)
=
c
o
t
α
−
(
c
o
t
α
−
1
c
o
t
α
)
=
c
o
t
α
−
2
(
c
o
t
2
α
−
1
2
c
o
t
α
)
=
c
o
t
α
−
2
(
c
o
s
2
α
−
sin
2
α
2
c
o
s
α
sin
α
)
=
c
o
t
α
−
2
c
o
t
2
α
=
R
.
H
.
S
.
of (1)
Therefore,
P
(
1
)
is true.
Step II: Assume it is true for
n
=
k
, then
P
(
k
)
:
t
a
n
α
+
2
t
a
n
2
α
+
2
2
t
a
n
2
2
α
+
.
.
.
.
+
2
k
−
1
t
a
n
(
2
k
−
1
α
)
=
c
o
t
α
−
2
k
c
o
t
(
2
k
α
)
Step III: For
n
=
k
+
1
,
P
(
k
+
1
)
:
t
a
n
α
+
2
t
a
n
2
α
+
2
2
t
a
n
2
2
α
+
.
.
.
+
2
k
−
1
t
a
n
(
2
k
−
1
α
)
+
2
k
t
a
n
(
2
k
α
)
=
c
o
t
α
−
2
k
+
1
c
o
t
(
2
k
+
1
α
)
L
.
H
.
S
.
=
t
a
n
α
+
2
t
a
n
2
α
+
2
2
t
a
n
2
2
α
+
.
.
.
.
+
2
k
−
1
t
a
n
(
2
k
−
1
α
)
+
2
k
t
a
n
(
2
k
α
)
=
c
o
t
α
−
2
k
c
o
t
(
2
k
α
)
+
2
k
t
a
n
(
2
k
α
)
(By assumption step)
=
c
o
t
α
−
2
k
(
c
o
t
(
2
k
α
)
−
t
a
n
(
2
k
α
)
)
=
c
o
t
α
−
2
k
⋅
2
(
c
o
t
2
(
2
k
α
)
−
1
2
c
o
t
(
2
k
α
)
)
=
c
o
t
α
−
2
k
+
1
⋅
c
o
t
(
2
⋅
2
k
α
)
=
c
o
t
α
−
2
k
+
1
⋅
c
o
t
(
2
k
+
1
α
)
=
R
.
H
.
S
.
This show that the result is true for
n
=
k
+
1
.
Hence by the principle of mathematical induction, the result is true for all n
∈
N.
If
p
is a prime number, then
n
p
−
n
is divisible by
p
for all
n
, where
Report Question
0%
n
∈
N
.
0%
n
is odd natural number.
0%
n
is even natural number.
0%
n
is not a composite number.
Explanation
n
p
−
n
is divisible by
p
for
n
=
1
Let,
k
p
−
k
=
p
λ
, where
k
∈
N
and
λ
∈
Z
Now,
(
k
+
1
)
p
−
(
k
+
1
)
=
k
p
+
p
C
1
k
p
−
1
+
.
.
.
.
.
.
.
.
.
+
p
C
p
−
1
−
k
(
k
p
−
k
)
+
(
p
C
1
k
p
−
1
+
.
.
.
+
p
C
p
−
1
k
)
=
p
λ
+
(
p
C
1
k
p
−
1
+
.
.
.
+
p
C
p
−
1
k
)
which is divisible by
p
Reason :
p
C
f
is always divisible by by p when
f
≠
p
,
0
because p is a prime number as the denominator will not have any factors or multiples of p.
So by mathematical induction,
n
p
−
n
is divisible by
p
for all
n
∈
N
For each
n
ϵ
N
, then
3
2
n
+
1
+
1
is divisible by -
Report Question
0%
2
0%
3
0%
7
0%
None of these
Explanation
3
2
n
+
1
+
1
Let
n
=
1
.
Then,
3
2
n
+
1
+
1
=
3
(
2
×
1
)
+
1
+
1
=
3
2
+
1
+
1
3
3
+
1
=
28
which is divisible by
2.
Let
n
=
2
.
Then,
3
2
n
+
1
+
1
3
5
+
1
243
+
1
=
244
which is divisible by
2.
For positive integer n,
3
n
<
n
!
when
Report Question
0%
n
≥
6
0%
n
>
7
0%
n
≥
7
0%
n
≤
7
Explanation
For positive integer
n
,
3
n
<
n
!
Let
n
=
1
3
1
<
1
!
=
3
<
1
which is not possible.
Let
n
=
2
3
2
<
2
!
=
9
<
2
which is not possible.
Let
n
=
6
3
6
<
6
!
=
729
<
720
which is not possible.
Let
n
=
7
3
7
<
7
!
=
2187
<
5040
which is possible.
∴
For positive integer n,
3
n
<
n
!
when
n
≥
7
∀
n
∈
N
,
1
+
1
√
2
+
1
√
3
+
.
.
.
.
.
.
+
1
√
n
is
Report Question
0%
√
n
0%
≤
√
n
0%
>
√
n
0%
none of these
Explanation
Let
P
(
n
)
=
1
+
1
√
2
+
1
√
3
+
.
.
.
.
.
.
+
1
√
n
....(1)
Put
n
=
2
in (1)
P
(
2
)
=
1
+
1
√
2
>
√
2
Put
n
=
3
in (1)
P
(
3
)
=
1
+
1
√
2
+
1
√
3
>
√
3
So, let us assume
P
(
n
)
=
1
+
1
√
2
+
1
√
3
+
.
.
.
.
.
.
+
1
√
n
>
√
n
So, we will check for
P
(
n
+
1
)
P
(
n
+
1
)
=
1
+
1
√
2
+
1
√
3
+
.
.
.
.
.
.
+
1
√
n
+
1
√
n
+
1
P
(
n
+
1
)
>
√
n
+
1
√
n
+
1
=
√
n
(
n
+
1
)
+
1
√
n
+
1
....(2)
Since,
√
n
(
n
+
1
)
>
n
⇒
√
n
(
n
+
1
)
+
1
>
n
+
1
⇒
√
n
(
n
+
1
)
+
1
√
n
+
1
>
n
+
1
√
n
+
1
⇒
√
n
(
n
+
1
)
+
1
√
n
+
1
>
√
n
+
1
....(3)
From (2) and (3), it follows that
1
+
1
√
2
+
1
√
3
+
.
.
.
.
.
.
+
1
√
n
+
1
√
n
+
1
>
√
n
+
1
Hence, by mathematical induction method, we can say that
1
+
1
√
2
+
1
√
3
+
.
.
.
.
.
.
+
1
√
n
>
√
n
for
n
∈
N
,
n
≥
2
Let
p
(
n
)
=
x
(
x
n
−
1
−
n
⋅
a
n
−
1
+
a
n
(
n
−
1
)
)
is divisible by
(
x
−
a
)
2
for
Report Question
0%
n
>
1
0%
n
>
2
0%
∀
n
∈
N
0%
None of these
Explanation
p
(
1
)
=
x
Hence it is not divisible by
(
x
−
a
)
2
.
p
(
2
)
=
x
(
x
−
2
a
+
a
2
)
=
x
2
−
2
a
x
+
a
2
x
=
(
x
−
a
)
2
−
a
2
+
a
2
(
x
)
=
(
x
−
a
)
2
+
a
2
(
x
−
1
)
...(i)
Hence
p
(
2
)
is also not divisible by
(
x
−
a
)
2
.
Now
{
1
,
2
}
falls in the set of first n natural numbers.
Hence the correct answer is
Option D.
If
10
n
+
3
⋅
4
n
+
2
+
λ
is exactly divisible by
9
for all
n
∈
N
, then the least positive integral value of
λ
is
Report Question
0%
5
0%
3
0%
7
0%
1
Explanation
Let
P
(
n
)
=
10
n
+
3
⋅
4
n
+
2
+
λ
Since
P
(
n
)
is divisible by
9
for all
n
∈
N
⇒
P
(
1
)
will be also divisible by 9
⇒
P
(
1
)
=
10
+
3
(
4
3
)
+
λ
=
202
+
λ
=
22
×
9
+
(
λ
+
4
)
Thus least positive integral value of
λ
for which
P
(
1
)
can be divisible by
9
is
5
Hence option 'A" is correct choice.
For all
n
∈
N
,
10
n
+
3.4
n
+
2
+
5
is divisible by
Report Question
0%
23
0%
3
0%
9
0%
207
Explanation
10
n
+
3.4
n
+
2
+
5
Put
n
=
1
10
+
3
(
64
)
+
5
=
207
which is divisible by
3
,
9
,
23
,
207
We will check for
n
=
2
100
+
768
+
5
=
873
which is divisible by
3
and
9
only.
Every number divisible by
9
is also divisible by
3
.
Now, let
P
(
n
)
is true
i.e.
10
n
+
3.4
n
+
2
+
5
is divisible by
9
⇒
10
n
+
3.4
n
+
2
+
5
=
9
λ
....(1)
Now, we will check for
P
(
n
+
1
)
Consider,
10
n
+
1
+
3.4
n
+
3
+
5
=
10
n
10
+
3.4
n
+
2
4
+
5
=
10
(
9
λ
−
3.4
n
+
2
−
5
)
+
3.4
n
+
2
4
+
5
(by (1))
=
90
λ
−
27.4
n
+
2
−
45
=
9
(
10
λ
−
3.4
n
+
2
−
5
)
which is divisible by 9.
Hence, by mathematical induction given expression is divisible by
3
,
9
.
If
4
n
C
2n
:
2
n
C
n
=
{
1.3
,
.5
,
.
.
.
(
4
n
−
1
)
}
:
{
1
,
3
,
5....
(
2
n
−
1
)
}
λ
,
t
h
e
n
λ
=
Report Question
0%
1
0%
2
0%
3
0%
none of these
Explanation
4
n
C
2
n
:
2
n
C
n
=
{
1
,
3
,
5
,
…
.
(
4
n
−
1
)
}
;
{
1
,
3
,
5
,
…
.
(
2
n
−
1
)
}
We have
2
n
!
n
!
=
2
n
(
2
n
−
1
)
(
2
n
−
1
)
…
.3
.2
.1
n
!
=
[
2.4.6
…
.
(
2
n
−
2
)
(
2
n
)
]
[
1.3.5
…
.
(
2
n
−
1
)
]
/
n
!
Writing the odd terms and even terms we get
2
n
!
n
!
=
2
n
[
1.2.3
…
.
(
n
−
1
)
.
n
]
[
1.3.5
…
.
(
2
n
−
1
)
]
/
n
!
=
2
n
n
!
[
1.3.5
…
.
(
2
n
−
1
)
.
n
]
n
!
=
[
1.3.5
…
.
(
2
n
−
1
)
]
Replacing
n
by
2
n
,
4
n
!
2
n
!
=
2
2
n
[
1.3.5
…
.
(
4
n
−
1
)
]
Now we have
4
n
C
2
n
:
2
n
C
n
=
4
n
!
2
n
!
2
n
!
×
n
!
n
!
2
n
!
=
4
n
!
2
n
!
.
(
n
!
2
n
!
)
2
Now
4
n
C
2
n
!
2
n
C
n
=
2
2
n
[
1.3.5
…
.
(
4
n
−
1
)
]
[
2
n
(
1.3.5
…
.
(
2
n
−
1
)
)
]
=
[
1.3.5
…
.
(
4
n
−
1
)
]
[
1.3.5
…
.
(
2
n
−
1
)
]
2
⇒
λ
=
2
If
2
83
+
k
is divisible by 127, then the smallest positive integral value of k is:
Report Question
0%
63
0%
31
0%
15
0%
64
Using principle of mathematics induct or for all
n
∈
N
:
1
+
2
+
3
+
.
.
.
.
+
n
<
1
8
(
2
n
+
1
)
2
Report Question
0%
True
0%
False
For all
n
∈
N
,
1
2
+
2
2
+
3
2
+
4
2
+
⋯
+
n
2
=
n
(
n
+
1
)
(
2
n
+
1
)
6
Report Question
0%
True
0%
False
1
1
2
+
1
2
2
+
3
3
2
+
.
.
.
.
.
.
+
1
n
2
≤
3
n
+
1
2
n
+
2
for every natural number
n
Report Question
0%
True
0%
False
For any natural number
n
,
x
n
y
n
is divisible by
x
y
,
where
x
and
y
are any
integers with
x
≠
y
.
Report Question
0%
True
0%
False
0:0:1
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3
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8
9
10
11
12
13
14
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16
17
18
19
20
21
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2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
0
Answered
1
Not Answered
20
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Correct : 0
Incorrect : 0
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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