Byju's Answer
Standard XII
Mathematics
Properties of Determinants
If $ D - k = ...
Question
If
D
k
=
∣
∣ ∣ ∣
∣
1
n
n
2
k
n
2
+
n
+
1
n
2
+
n
2
k
−
1
n
2
n
2
+
n
+
1
∣
∣ ∣ ∣
∣
and
n
∑
k
=
1
D
k
=
56
,
where
n
∈
N
,
then
n
is equal to
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Solution
It is given that
n
∑
k
=
1
D
k
=
56
where,
D
k
=
∣
∣ ∣ ∣
∣
1
n
n
2
k
n
2
+
n
+
1
n
2
+
n
2
k
−
1
n
2
n
2
+
n
+
1
∣
∣ ∣ ∣
∣
⇒
∣
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
∣
n
∑
k
=
1
1
n
n
n
∑
k
=
1
2
k
n
2
+
n
+
1
n
2
+
n
n
∑
k
=
1
(
2
k
−
1
)
n
2
n
2
+
n
+
1
∣
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
∣
=
56
n
∑
k
=
1
1
=
n
n
∑
k
=
1
2
k
=
2
(
1
+
2
+
⋯
+
n
)
=
2
n
(
n
+
1
)
2
=
n
(
n
+
1
)
n
∑
k
=
1
(
2
k
−
1
)
=
n
∑
k
=
1
2
k
−
n
∑
k
=
1
1
=
n
(
n
+
1
)
−
n
=
n
2
⇒
∣
∣ ∣ ∣
∣
n
n
n
n
(
n
+
1
)
n
2
+
n
+
1
n
2
+
n
n
2
n
2
n
2
+
n
+
1
∣
∣ ∣ ∣
∣
=
56
C
2
→
C
2
−
C
1
,
C
3
→
C
3
−
C
1
∣
∣ ∣ ∣
∣
n
0
0
n
(
n
+
1
)
1
0
n
2
0
n
+
1
∣
∣ ∣ ∣
∣
=
56
⇒
n
(
n
+
1
)
=
56
⇒
n
2
+
n
−
56
=
0
⇒
n
2
+
8
n
−
7
n
−
56
=
0
⇒
(
n
+
8
)
(
n
−
7
)
=
0
⇒
n
=
7
,
n
=
−
8
(Not Possible)
∴
n
=
7
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Similar questions
Q.
If
D
k
=
∣
∣ ∣ ∣
∣
1
n
n
2
k
n
2
+
n
+
1
n
2
+
n
2
k
−
1
n
2
n
2
+
n
+
1
∣
∣ ∣ ∣
∣
and
∑
n
k
=
1
D
k
=
56
, then n equals
Q.
If
D
k
=
∣
∣ ∣ ∣
∣
1
n
n
2
k
n
2
+
n
+
1
n
2
+
n
2
k
−
1
n
2
n
2
+
n
+
1
∣
∣ ∣ ∣
∣
and
∑
n
k
=
1
D
k
=
56
, then
n
equals ?
Q.
If
D
k
=
∣
∣ ∣ ∣
∣
1
n
n
2
k
n
2
+
n
+
1
n
2
+
n
2
k
−
1
n
2
n
2
+
n
+
1
∣
∣ ∣ ∣
∣
and
n
∑
k
=
1
D
k
=
56
,
where
n
∈
N
,
then
n
is equal to
Q.
For a fixed positive integer n, if
D
=
∣
∣ ∣ ∣
∣
n
!
(
n
+
1
)
!
(
n
+
2
)
!
(
n
+
1
)
!
(
n
+
2
)
!
(
n
+
3
)
!
(
n
+
2
)
!
(
n
+
3
)
!
(
n
+
4
)
!
∣
∣ ∣ ∣
∣
, then
[
D
(
n
!
)
3
−
4
]
is
___
Q.
For a fixed positive integer n, if
D
=
∣
∣ ∣ ∣
∣
n
!
(
n
+
1
)
!
(
n
+
2
)
!
(
n
+
1
)
!
(
n
+
2
)
!
(
n
+
3
)
!
(
n
+
2
)
!
(
n
+
3
)
!
(
n
+
4
)
!
∣
∣ ∣ ∣
∣
then show
[
D
(
n
!
)
3
−
4
]
is divisible by n.
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