<s>[INST] Partial sum of $k>2$ consecutive positive integers has no solution in prime Progress:

Let $a_1,...,a_k$ a sequence of $k$ consecutive positive integers. If $k>2$, show that no partial sum of the sequence $a_1+...+a_k$ has a solution in prime.
I made a heuristic attempt and found a pattern in the progression below:

$$x$$ $$x+(x+1)=2x+1$$ $$x+(x+1)+(x+2)=3x+3$$ $$x+(x+1)+(x+2)+(x+3)=4x+6$$ $$x+(x+1)+(x+2)+(x+3)+(x+4)=5x+10$$ $$x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=6x+15$$ $$x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)=7x+21$$ $$...$$

Which clearly shows that after $2$nd line each right-hand side value tend to have a common factor greater than $1$ and implies they cannot be prime. Although, technically it's not a proof, so I would appreciate any elementary proof on this matter.

Regards
 [/INST] The sum $a+(a+1)+\dots + (a+k-1)$ is equal to $\frac{k(2a+k-1)}{2}$.
If $k>2$ then both $k$ and $(2a+k-1)>2$. So your number is the product of two numbers that are larger than $1$.
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