An algebra problem by Abu Bakar Khan

We know that,

$i^{4n+1}=i^{1}=i \\ i^{4n+2}=i^{2}=-1 \\ i^{4n+3}=i^{3}=-i \\ i^{4n}=i^{4}=1$

Now,

$\displaystyle \sum_{r=1}^{25}{i^{r!}} = i^{1!} + i^{2!} + i^{3!} + i^{4!} + \cdots + i^{25!}$

$\because 4|(4+x)!$ ,where $x\in\mathbb{Z}$ and $x\ge 0$ .

So,

$i^{4!}=i^{5!}=i^{6!}=\cdots=i^{25!}=1$

Hence,

$~~~~~~~\displaystyle \sum_{r=1}^{25}{i^{r!}} = i + i^{2} + i^{6} + \underbrace{1 + 1 + \cdots + 1}_{22~~ \text{times}} \\ \Longrightarrow \displaystyle \sum_{r=1}^{25}{i^{r!}} = i + (-1) + (-1) + 22 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \color{#302B94}{\because i^{6}=i^{4\times1+2}=i^{2}=-1} \\ \Longrightarrow \displaystyle \sum_{r=1}^{25}{i^{r!}} = \boxed {i + 20}$