# Math Help - sequence problem

1. ## sequence problem

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!

2. Originally Posted by Not2l8
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!
Sorry I see no obvious pattern.

3. Hello, Not2l8!

What a strangely-written problem!
And is there a typo?

Solve for $k\!:\;\;1 + 14(k+1) + 12(1+2+3+\hdots k) \;=\; {\color{red}k}$

We have: . $1 + 14k + 14 + 12\cdot\frac{k(k+1)}{2} \;=\;k \quad\Rightarrow\quad 1 + 14k + 14 + 6k^2 + 6k \;=\;k$

. . $6k^2 + 19k + 15 \:=\:0 \quad\Rightarrow\quad (2k + 3)(3k + 5) \:=\:0$

Therefore: . $k \;=\;-\frac{3}{2},\:-\frac{5}{3}$

4. 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!