1. The sequence is given by $a_{n}=\frac{c^{2}+n-2}{2}, n \in \mathbb{N}$.

a) Determine the sum $S_{n}$ of the first $n$ terms.
b) Prove that the numerator of the sum $S_{c}$ is divisible by 25 if $c=10 k+1$, where $k$ is a natural number.
Solution. a) We have:

$$
S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} \frac{c^{2}+k-2}{2}=n \frac{c^{2}-2}{2}+\frac{1}{2}(1+2+\ldots+n)=\frac{n}{4}\left(2 c^{2}+n-3\right)
$$

b) If $c=10 k+1$, then the numerator of $S_{c}$ is equal to

$$
(10 k+1)\left(2(10 k+1)^{2}+10 k+1-3\right)=25 k(10 k+1)(8 k+2)
$$