Another Summation Question

[Mckanic]

[MATH]\sum_{k=1}^4(xk/(5+\sum_{i=1}^kx(6-i))[/MATH] = z

= x1/(5+5x+4x+3x+2x) + x2/(5+5x+4x+3x) + x3/(5+5x+4x) + x4/(5+5x) = z

now i need 4 equations

[MATH]j^0\sum_{k=1}^4(xk/(5+\sum_{i=1}^kx(6-i))[/MATH] = z

j=1....4

= 1*(x1/(5+5x+4x+3x+2x) + x2/(5+5x+4x+3x) + x3/(5+5x+4x) + x4/(5+5x))= z
= 1*(x1/(5+5x+4x+3x+2x) + x2/(5+5x+4x+3x) + x3/(5+5x+4x) + x4/(5+5x))= z
= 1*(x1/(5+5x+4x+3x+2x) + x2/(5+5x+4x+3x) + x3/(5+5x+4x) + x4/(5+5x))= z
= 1*(x1/(5+5x+4x+3x+2x) + x2/(5+5x+4x+3x) + x3/(5+5x+4x) + x4/(5+5x))= z

Assuming I am correct to this part

Now I need to reduce the denominator and increase numerator

=(x1/(5+5x+4x+3x+2x) + x2/(5+5x+4x+3x) + x3/(5+5x+4x) + x4/(5+5x)= z
=(x1+x2/(5+5x+4x+3x) + x2+x3/(5+5x+4x) + x3+x4/(5+5x) = z
=(x1+x2+x3/(5+5x+4x) + x2+x3+x4/(5+5x) = z
=(x1+x2+x3+x4/(5+5x) = z

Looks like numerator would need a double summation

[MATH]j^0(\sum_{i=1}^4\sum_{k=1}^ixk)/(5+\sum_{i=1}^kx(6-i)[/MATH] = z

But this doesn't seem correct.

Denominator still has me stuck at the moment.

 

[Dr.Peterson]

[MATH]\sum_{k=1}^4(xk/(5+\sum_{i=1}^kx(6-i))[/MATH] = z

= x1/(5+5x+4x+3x+2x) + x2/(5+5x+4x+3x) + x3/(5+5x+4x) + x4/(5+5x) = z

This doesn't make sense. I suppose you meant xk to be a subscript; but in any case, your first term has k=1, so the summation within it should have only one term after the 5, not 4. Please correct this before asking questions based on it.

Also, it will help if you explain your goal. Where does this come from?