Question

Let m be a positive integer and ∆_r

=$\left| \begin{matrix} 2r - 1 & mC_{r} & 1 \\ m^{2} - 1 & 2^{m} & m + 1 \\ \sin^{2}(m^{2}) & \sin^{2}(m) & \sin^{2}(m + 1) \end{matrix} \right|$= (0 ≤ r ≤ m) then the value of $\sum_{r = 0}^{m}\Delta_{r}$is given by –

• A. 0
• B. m² – 1
• C. 2^m
• D. 2^m sin² (2^m)

Answer 1

Answer: (A)

0

Answer 2

$\sum_{r = 0}^{m}{(2r - 1)}$= sum of (m + 1) terms of an A.P.

whose first term is –1, d = 2 ∴ S = m² – 1

$\sum_{r = 0}^{m}{mC_{r}}$= sum of binomial coefficient = 2^m $\sum_{}^{}1$= m + 1

∴ R₁ and R₂ will be identical and hence ∆ = 0.