Multiply the sum and write result in summation notation.?

Question

Multiply out (∑x_i)2, writing the result in summation notation. The sum goes from i=1 to n.

sorry, when i copied the question from word format it made the 2 become un-squared!

Yes, the whole thing is squared (SUM x_i)^2!

Answer 1

i guess u mean all squared?
= (x_1 + x_2 + ... + x_n ) ^2 =
(x_1 + x_2 + ... + x_n )(x_1 + x_2 + ... + x_n )
= (∑x_i x_j), where i=1 to n, j=1 to n.

right?

Answer 2

we hit upon the inverse of the matrix by technique of the co-aspect technique and For the first row: the cofactor of two is a million x 0 - 4 x (-a million) = 0 + 4 = 4 the cofactor of four is -a million x 0 - a million x (-a million) = 0 + a million = a million the cofactor of one million is -a million x 4 - a million x a million = -4 -a million = -5 For the 2d row: the cofactor of -a million is 4 x 0 - 4 x a million = 0 -4 = -4 the cofactor of one million is two x 0 - a million x (a million) = 0 - a million = -a million the cofactor of -a million is two x 4 - a million x a million = 8 -4 = 4 For the third row: the cofactor of one million is 4 x-a million - a million x (a million) = -4 -a million = -5 the cofactor of four is two x -a million - a million x (-a million) =-2+ a million = -a million the cofactor of 0 is two x a million - -a million x 4 = 2 +4 = 6 So the cofactor matrix received is 4 a million -5 -4 -a million 4 -5 -a million 6 replacing the signs and indications + - + - + - + - + we get 4 -a million -5 4 -a million -4 -5 a million 6 Transposing this matrix and dividing by technique of determinant of A we get ( the cost of determinant A = 2(1x0 - 4 x -a million) - 4(-1x0 - 1x -a million) + a million(-a million x 4- 1x a million) = 2 (4) - 4(a million) + a million(-5) = 8 -4 -5 = -a million -4 -4 5 a million a million -a million 5 4 -6 therefore our answer