Alan

As follows:

The left hand side of the expression is 1 if (i) $x^2 -5x+2=0$   or  (ii) $x^2 - 4x + 2 = 1$

These are quadratics that can be solved using the standard quadratic formula.  They will give values of x in the form $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ 

Adding these two values will cancel the discriminant part leaving you with a sum of $\frac{-b}{a}$.  

Do this for each quadratic and then add the results.

Also if $x^2 - 4x + 2 = -1$ and $x^2 -5x+2$ is even then the RHS will be 1  (This will give you two more x values).

I suspect mingus means $\frac{-47+69i}{7+6i}$ though only he/she will know for certain!

I interpreted this question as follows:

If x is between 1 and 10 then f(x) = -1

If x is between 11 and 1000 then f(x) = -2

Hence the sum is 10*(-1) + 990*(-2) =...

Like this:

The second expression is simply a rearrangement of the first:

$\frac{6x^3+9x^2+4x+7}{2x+3} = \frac{3x^2(2x+3)+4x+6+1}{2x+3}= \frac{3x^2(2x+3)+2(2x+3)+1}{2x+3}\\ = \frac{3x^2(2x+3)}{2x+3}+\frac{2(2x+3)}{2x+3}+\frac{1}{2x+3} = 3x^2+2+\frac{1}{2x+3}$

Sum to infinity =  $\frac{18}{1-r}$, so $\frac{18}{1-r}=20$

Sum to n'th term = $\frac{18(1-r^n)}{1-r}$, so sum of subsequent terms = $20 - \frac{18(1-r^n)}{1-r}$

n'th term = $18r^{n-1}$

Can you complete the problem from here?

Answered here:  https://web2.0calc.com/questions/help-please_29808#2

This should help.