What is the inverse of the following (n x n)-matrix?
$$ \begin{bmatrix} 2 &-1 &0 &0 &... &0 &0 \\ -1 &2 &-1 &0 &... &0 &0 \\ 0 &-1 &2 &-1 &... &0 &0 \\ 0 &0 &-1 & 2 &... &0 &0 \\ ... &... &... &... &... &... &... \\ 0 &0 &0 &0 &... &2 &-1 \\ 0 &0 &0 &0 &... &-1 &2 \end{bmatrix} $$
Using software to calculate the answer for $n$ up to $7$ suggests the following answer: $$a_{i,j}=\cases{\frac{i(n+1-j)}{n+1}&if $i\le j$\cr \frac{j(n+1-i)}{n+1}&if $i\ge j$,\cr}$$ which is then easy to prove.