If Δ r = 2 r - 1 m C r 1 m 2 - 1 2 m m + 1 sin 2 ⁡ m 2 sin 2 ⁡ m sin 2 ⁡ m + 1 , then the value of ∑ r = 0 m Δ is

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Correct answer: D. 0

Δ r = 2 r - 1 m C r 1 m 2 - 1 2 m m + 1 s i n 2 m 2 sin 2 ⁡ m sin 2 ⁡ m + 1 ∴ ∑ r = 0 m Δ r = ∑ r = 0 m 2 r - 1 ∑ r = 0 m m C r ∑ r = 0 m 1 m 2 - 1 2 m m + 1 sin 2 ⁡ m 2 sin 2 ⁡ m sin 2 ⁡ m + 1 = m 2 - 1 2 m m + 1 m 2 - 1 2 m m + 1 sin 2 ⁡ m 2 sin 2 ⁡ m sin 2 ⁡ m + 1 = 0 ( ∵ two rows are identical)

If Δ r = 2 r - 1 m C r 1 m 2 - 1 2 m m + 1 sin 2 ⁡ m 2 sin 2 ⁡ m sin 2 ⁡ m + 1 , then the value of ∑ r = 0 m Δ is

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