Given $B=\{e_1,\dots,e_n\}$ a basis of a K-vector space $V$, and $f$ the only endomorphism such that
$f(e_i)=\begin{cases} e_1-e_2, & \mbox{if } i=1 \\ -e_{i-1}+e_i-e_{i+1}, & \mbox{if } 2 \le i \le n-1 \\ -e_{n-1}+e_n, & \mbox{if } i=n \end{cases} $
I have to prove that f is invertible iff $n \equiv 0,1\mod 3$.
If that's the case, show that
$f^{-1}(e_1)=\begin{cases} e_1+\sum_{k=1}^m (-1)^k(e_{3k}+e_{3k+1}) & \mbox{if } n=3m+1 \\ \sum_{k=1}^m (-1)^k(e_{3k-1}+e_{3k}) & \mbox{if } 2 \le i \le n-1 \end{cases}$
I've already proved that f is bijective iff $n \equiv 0,1 \mod3$, and now I'm proving the second part, with the case $n=3m+1$. Here's what I do.
Supose that $n=3m+1$. Using the definition of f:
$f(e_1+\sum_{k=1}^m (-1)^k(e_{3k}+e_{3k+1}))=\\ =f(e_1)+\sum_{k=1}^{m-1}(-1)^kf(e_{3k}+e_{3k+1})+(-1)^mf(e_{3m})+(-1)^mf(e_{3m+1})\\ =e_1-e_2+\sum_{k=1}^{m-1}(-1)^k(-e_{3k_1}-e_{3k+2})+(-1)^m(-e_{3m-1})$
From here I don't know how to continue.
Note that a lot of terms cancel due to the alternating sign: $$ \sum_{k=1}^{m-1} (-1)^k( -e_{3k-1} - e_{3k+2} ) = e_2 - (-1)^{m-1} e_{3m-1} = e_2 +(-1)^me_{3m-1}. $$ Then $$ f(e_1 + \sum(\dots)) = e_1. $$