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Question: If m is a positive integer D<sub>r</sub>=\(\left| \begin{matrix} 2r - 1 & mC_{r} & 1 \\ m^{2} - 1 &...

If m is a positive integer

Dr=∣2r−1mCr1m2−12mm+1sin⁡2(m2)sin⁡2(m)sin⁡2(m+1)∣\left| \begin{matrix} 2r - 1 & mC_{r} & 1 \\ m^{2} - 1 & 2^{m} & m + 1 \\ \sin^{2}(m^{2}) & \sin^{2}(m) & \sin^{2}(m + 1) \end{matrix} \right|Then find the value of ∑r=0mΔr\sum_{r = 0}^{m}\Delta_{r}

A

0

B

1

C

2

D

3

Answer

0

Explanation

Solution

∑r=0mΔr\sum_{r = 0}^{m}\Delta_{r}=∣∑r=0m(2r−1)∑r=0mmCr1m2−12mm+1sin⁡2(m2)sin⁡2(m)sin⁡2(m+1)∣\left| \begin{matrix} \sum_{r = 0}^{m}{(2r - 1)} & \sum_{r = 0}^{m}{mC_{r}} & 1 \\ m^{2} - 1 & 2^{m} & m + 1 \\ \sin^{2}(m^{2}) & \sin^{2}(m) & \sin^{2}(m + 1) \end{matrix} \right|

=∣m2−12mm+1m2−12mm+1sin⁡2(m2)sin⁡2(m)sin⁡2(m+1)∣\left| \begin{matrix} 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) \end{matrix} \right| = 0