Consider the sequence defined by $a_k =\dfrac{1}{k^2+k}$ for $k\geq 1$. Given that $a_m+a_{m+1}+\cdots+a_{n-1}=\dfrac{1}{29}$, for positive integers $m$ and $n$ with $m<n$, find $m+n$.
$\dfrac{1}{k^2+k}=\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$. Thus,
$a_m+a_{m+1}+\cdots +a_{n-1}=\dfrac{1}{m}-\dfrac{1}{m+1}+\dfrac{1}{m+1}-\dfrac{1}{m+2}+\cdots +\dfrac{1}{n-1}-\dfrac{1}{n}=\dfrac{1}{m}-\dfrac{1}{n}$
Which means that
$\dfrac{n-m}{mn}=\dfrac{1}{29}$
Since we need a factor of 29 in the denominator, we let $n=29t$.* Substituting, we get
$29t-m=mt$
so
$\frac{29t}{t+1} = m$
Since $m$ is an integer, $t+1 = 29$, so $t=28$. It quickly follows that $n=29(28)$ and $m=28$, so $m+n = 30(28) = \fbox{840}$.

*If $m=29t$, a similar argument to the one above implies $m=29(28)$ and $n=28$, which implies $m>n$. This is impossible since $n-m>0$.