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Question
Mathematics
If Δ r = |2r-1&mCr&1 m2-1&2m&m+1 sin2(m2)&sin2(m)&sin2(m+1)| then value of ∑ limitsmr = 0 Δ r is
Q. If
Δ
r
=
∣
∣
2
r
−
1
m
2
−
1
s
i
n
2
(
m
2
)
m
C
r
2
m
s
i
n
2
(
m
)
1
m
+
1
s
i
n
2
(
m
+
1
)
∣
∣
then value of
r
=
0
∑
m
Δ
r
is
2434
217
Determinants
Report Error
A
0
72%
B
4
10%
C
3
7%
D
1
11%
Solution:
Δ
r
=
∣
∣
2
r
−
1
m
2
−
1
s
i
n
2
(
m
2
)
m
C
r
2
m
s
i
n
2
(
m
)
1
m
+
1
s
i
n
2
(
m
+
1
)
∣
∣
⇒
r
=
0
∑
m
Δ
r
=
∣
∣
r
=
0
∑
m
(
2
r
−
1
)
m
2
−
1
s
i
n
2
(
m
2
)
r
=
0
∑
m
m
C
r
2
m
s
i
n
2
(
m
)
r
=
0
∑
m
1
m
+
1
s
i
n
2
(
m
+
1
)
∣
∣
=
∣
∣
m
2
−
1
m
2
−
1
s
i
n
2
(
m
2
)
2
m
2
m
s
i
n
2
(
m
)
m
+
1
m
+
1
s
i
n
2
(
m
+
1
)
∣
∣
=
0