Question

What is the value of r = 1Σ^n f(x) if  f(r) = \(\begin{vmatrix}2r & x & n(n + 1) \\(6r^2 – 1) & y & n^2 (2n + 3) \\(4r^3 – 2nr) & z & n^3(n + 1) \end {vmatrix}\)  where n € N?

(a) 1

(b) -1

(c) 0

(d) 2

I got this question at a job interview.

My question is taken from Application of Determinants topic in portion Determinants of Mathematics – Class 12

Answer 1

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