summation - Proofing $sum^n_{k=0}(-1)^k{nchoose k}(a_1+kd)=0$

I came across this following problem in an old math textbook:

Proof that $$\sum^n_{k=0}(-1)^k{n\choose k}a_{k+1}=0$$ where $n\geq 2$, ${n \choose k}$ is the binomial coefficient and $(a_k)$ is an arithmetic sequence of the form $a_k=a_1+(k-1)d$ with $a_1,d\in \mathbb{R}$.

My first try was to proof it with induction, but it wasn't useful. Furthermore I tried "seperating" the sum into odd and even indices by writing a sum as $$\sum^n_{k=0}z_k=\sum^\alpha_{k=0}z_{2k+1}+\sum^\beta_{k=0}z_{2k}$$ where $\alpha$ and $\beta$ depend on $n$. It would be a pleasure if you helped me.

Edit: I forgot to mention that derivatives weren't introduced yet in this textbook so I'm looking for ways without using them.

Answer

We need to prove $\sum_{k=0}^n(-1)^k\binom{n}{k}x^k,\,\sum_{k=0}^n(-1)^k\binom{n}{k}kx^{k-1}$ each vanish at $x=1$. These functions are respectively $(1-x)^n$ and its derivative, $-n(1-x)^{n-1}$. We can also avoid calculus in the second problem by noting $$\sum_{k=0}^n(-1)^k\binom{n}{k}k=\sum_{k=1}^n(-1)^k\binom{n}{k}k=n\sum_{k=1}^n(-1)^k\binom{n-1}{k-1}=n\sum_{j=0}^{n-1}(-1)^k\binom{n-1}{j},$$thus reducing to the first problem.