1.

What is the value of r = 1Σ^n f(x) iff(r) = \(\begin{vmatrix}2r & x & n(n + 1) \\(6r^2 – 1) & y & n^2 (2n + 3) \\(4r^3 – 2nr) & z & n^3(n + 1) \end {vmatrix}\)where n € N?(a) 1(b) -1(c) 0(d) 2I got this question at a job interview.My question is taken from Application of Determinants topic in portion Determinants of Mathematics – Class 12

Answer»

Right CHOICE is (c) 0

The best EXPLANATION: The given determinant is f(r) = \(\begin{vmatrix}2R & x & N(n + 1) \\(6r^2 – 1) & y & n^2(2n + 3) \\(4r^3 – 2nr) & z & n^3(n + 1) \end {vmatrix}\)

Now, r = 1Σ^n (2r) = 2[(n(n + 1))/2]……….(1)

= n^2 + n

r = 1Σ^n(6r^2 – 1) = 6[((n + 1)(2n + 1))/6] – n……….(2)

= n(2n^2 + 2n + n + 1) – n

= 2n^3 + 2n^2 + n^2 + n – n

= 2n^3 + 3n^2

= r = 1Σ^n(4r^3 – 2nr) = n^3 (n + 1)……….(3)

From (1), (2) and (3) we get

r = 1Σ^n f(x) = 0


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