Alan

Like this (assuming my simplification of the numbers is correct):

I'll leave you to solve the above to find the actual value of Cfixed.

Hmm!  Check the left-hand side of 15f = 5m

As follows:

The following is one possibility:

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.