Let `"Delta"_r=|r x(n(n+1))/2 2r-1y n^2 3r-2z(n(3n-1))/2|dot` Show that `sum_(r=1)^n"Delta"_r=0`

Options

A. xyz
B. n xyz
C. 0
D. none of these

Answer

C

Explanation

We know that `|(a_(1),x,p),(a_(2),y,q),(a_(3),z,r)|+ |(b_(1),x,p),(b_(2),y,q),(b_(3),z,r)| + |(c_(1),x,p),(c_(2),y,q),(c_(3),z,r)| = |(a_(1) + b_(1) + c_(1),x,p),(a_(2) + b_(2) + c_(2),y,q),(a_(3) + b_(3) + c_(2),z,r)|`
i.e., the sum of the determinants having all columns (or rows) identical except a specific column (or row), say first, is a determinant whose first column (or row) is the sum of the corresponding elements of first columns of various determinants and the remaining columns (or row) remain same.
Using this property, we have
`underset(r=1)overset(n) sum Delta_(r) = |(underset(r=1)overset(n)sumr,x,(n(n+1))/(2)),(underset(r=1)overset(n)sum (2r -1),y,n^(2)),(underset(r=1)overset(n)sum (3r -2),z,(n(3n-1))/(2))|`
Now, `underset(r=1)overset(n)sum r = 1 + 2 + ...+ n = (n(n+1))/(2)`
`underset(r=1)overset(n)sum (2r -1)= 1 + 3 + 5 + ...+ (2n -1) = (n)/(2) {1 + (2n -1)} = n^(2)` and
`underset(r=1)overset(n)sum (3r -2) = 1 + 4 +7 +...+ (3n-2)`
`rArr underset(r=1)overset(n)sum (3r -2) = (n)/(2) [1 + (3n-2)] = (n(3n-1))/(2)`
`:. underset(r =1)overset(n)sum Delta_(r) = |((n(n+1))/(2),x,(n(n+1))/(2)),(n^(2),y,n^(2)),((n(3n-1))/(2),z,(n(3n-1))/(2))| = 0 " " [( :' C_(1) and C_(3)),("are identical")]`