Find a if $\sum_{K=1}^n f(a+K)=2n(33+n)$

Question

Let $f(x)$ be a function such that $$f(x+y)=f(x)+f(y)\forall x,y\in N$$ and $f(1)=4$. If $$\sum_{K=1}^n f(a+K)=2n(33+n)$$ then find $a$.

So I first calculated $$\sum_{K=1}^n f(a+K)= 2((a+n)(a+n+1)-a(a+1))$$ and equated it to the $RHS$. But the answer is not in terms of $n$. And you can see that my answer is in terms of $n$ which is of courses wrong since $a$ will surely be independent of$n$. Please help.

Answer 1

Let's compute again the sum, you know that $f(a)=f(a-1)+f(1)$ and inductively $f(a)=af(1)=4a.$ Same for $f(K)=4K$ and then $$\sum _{K = 1}^nf(a+K)=\sum _{K = 1}^nf(a)+f(K)=\sum _{K = 1}^n4a+4K=4(na+\frac{n(n+1)}{2})=4n(a+\frac{n+1}{2})=2n(2a+n+1),$$ hence

Can you conclude?

Answer 2

What you wrote is very correct. I don't understand your worries.
We have $$\forall n \geq 1, \sum_{K = 1}^{n} f(a +K) = 2 ((a +n) (a +n +1)- a (a+1)) = 2 n (n +2 a +1)$$ and hence, $$\forall n \geq 1, \sum_{K = 1}^{n} f(a +K) = 2 n (33 +n) \iff 2 a +1 = 33 \iff a = 16$$