1+(K+1)(14)+12(1+2+3+....(K-1)+K)=X
Solve for K?
Is this solvable? as I said before haven't done maths for years and this is making my head spin. If any one could help that would be great!
Hello, Not2l8!
What a strangely-written problem!
And is there a typo?
Solve for $\displaystyle k\!:\;\;1 + 14(k+1) + 12(1+2+3+\hdots k) \;=\; {\color{red}k} $
We have: .$\displaystyle 1 + 14k + 14 + 12\cdot\frac{k(k+1)}{2} \;=\;k \quad\Rightarrow\quad 1 + 14k + 14 + 6k^2 + 6k \;=\;k $
. . $\displaystyle 6k^2 + 19k + 15 \:=\:0 \quad\Rightarrow\quad (2k + 3)(3k + 5) \:=\:0 $
Therefore: .$\displaystyle k \;=\;-\frac{3}{2},\:-\frac{5}{3}$
Hi Soroban, thanks very much for your response.
Sorry I'm not really a maths guy hence the strangely written problem.
I think your answer was the one I was looking for but I was trying to solve for X at the end not K.
1+14(K+1)+12(1+2+3....+K)=X
And I think by your result I can see
6k^2+20K+15=X
which is awesome! That's what I was looking for thanks!