Mathematics

Mathematics

Problem

Let mm be a positive integer and Δr=∣2r−1mCr1m2−12mm+1sin⁡2(m2)sin⁡2(m)sin⁡2(m+1)∣(0≤r≤m).\Delta_r = \begin{vmatrix} 2r-1 & {}^m C_r & 1 \\ m^2-1 & 2^m & m+1 \\ \sin^2(m^2) & \sin^2(m) & \sin^2(m+1) \end{vmatrix} \quad (0 \leq r \leq m). Then the value of ∑r=0mΔr\sum_{r=0}^m \Delta_r is given by (a) 0 (b) m2−1m^2-1 (c) 2m2^m (d) 2msin⁡2(2m)2^m \sin^2(2^m) (a) 0 (b) m2−1m^2-1 (c) 2m2^m (d) 2msin⁡2(2m)2^m \sin^2(2^m)

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