Prove that $u_n$ is arithmetic sequence if $\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+\cdots+\frac{1}{u_{n-1}u_n}=\frac{n-1}{u_1u_n}$ Let $(u_n)$ be a sequence $u_i\neq0$ and
$$\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+\cdots+\frac{1}{u_{n-1}u_n}=\frac{n-1}{u_1u_n}$$ for all $n\geq3$ 
Prove that the sequence $(u_n)$ is arithmetic sequence
 A: We have
$$\rm s_n:=\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+....+\frac{1}{u_{n-1}u_n}=\frac{n-1}{u_1u_n}.$$
Thus,
$$\rm s_{n+1}-s_n=\frac{1}{u_n u_{n+1}}=\frac{1}{u_1}\left(\frac{n}{u_{n+1}}-\frac{n-1}{u_n}\right).$$
Multiply both sides by $\rm u_1u_nu_{n+1}$, negate, add $\rm u_{n+1}$ to both sides,
$$\rm u_{n+1}-u_1=n(u_{n+1}-u_n).$$
Suppose, as our induction hypothesis, that $\rm u_k=a_1+d(k-1)$ for $\rm k=1,2,\cdots,n$ and some $\rm d$. Then the equation above is linear in $\rm u_{n+1}$, thus has a unique solution, and plugging in $\rm u_{n+1}=a_1+nd$ clearly works so it must be precisely that solution. The base case $\rm a_1=a_1$ is clear and examining the next case $\rm n=2$ tells us $\rm d=a_2-a_1$.
A: There are cleverer ways of doing it, but here’s how you might approach it quasi-experimentally and arrive at a workable idea.
Start by looking at $n=3$: you have $$\frac1{u_1u_2}+\frac1{u_2u_3}=\frac2{u_1u_3}\;,$$ so $$\frac{u_1+u_3}{u_1u_2u_3}=\frac2{u_1u_3}\;,\tag{1}$$ and therefore $u_1+u_3=2u_2$. This says that $u_2$ is the mean of $u_1$ and $u_3$, so it’s midway between $u_1$ and $u_3$, and the first three terms therefore do indeed form an arithmetic progression. Let $d=u_2-u_1$.
What happens when $n=4$? You have $$\frac1{u_1u_2}+\frac1{u_2u_3}+\frac1{u_3u_4}=\frac3{u_1u_4}\;,$$ so $$\frac{u_3u_4+u_1u_4+u_1u_2}{u_1u_2u_3u_4}=\frac{3u_2u_3}{u_1u_2u_3u_4}\;,\tag{2}$$ and $$u_1u_2+u_3u_4+u_1u_4=3u_2u_3\;.\tag{3}$$ This implies that $$\begin{align*}
u_4&=\frac{3u_2u_3-u_1u_2}{u_1+u_3}\tag{4}\\
&=\frac{u_2(3u_3-u_1)}{u_1+u_3}\\
&=\frac{(u_1+d)(2u_1+6d)}{2u_1+2d}\\\\
&=u_1+3d\;,
\end{align*}$$
exactly what’s needed to keep it in arithmetic progression with the first three terms.
One more: $n=5$. The same kind of calculation leads to the equation $$u_5=\frac{4u_2u_3u_4-u_1u_2u_3}{u_1u_2+u_3u_4+u_1u_4}\;.\tag{5}$$ We already know from $(2)$ that the denominator of $(4)$ is $3u_2u_3$, so $$u_5=\frac{u_2u_3(4u_4-u_1)}{3u_2u_3}=\frac{3u_1+12d}3=u_1+4d\;,$$ as desired.
Now observe that the denominator in $(4)$ is the numerator in $(1)$, and the denominator in $(5)$ is the numerator in $(2)$. This should give you enough clues to pursue a proof by induction on $n$.
A: You need two key features of an arithmetic progression:
$$a_n = a_1 +d(n-1)$$
and 
$$a_n-a_{n-1}=d$$ (which is a consequence of the previous one).
Thus
$$\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+....+\frac{1}{u_{n-1}u_n}=\frac{n-1}{u_1u_n}$$
$$d\left(\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+....+\frac{1}{u_{n-
1}u_n}\right)=\frac{d(n-1)}{u_1u_n}$$
Now sum $\dfrac{u_1}{u_1 u_n}$ to get
$$\frac{u_1}{u_1 u_n} + d\left(\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}
{u_3u_4}+....+\frac{1}{u_{n-1}u_n}\right)=\frac{u_1+d(n-1)}{u_1u_n}$$
$$\frac{1}{u_n} + \frac{d}{u_1u_2}+\frac{d}{u_2u_3}+\frac{d}{u_3u_4}+....+\frac{d}{u_{n-1}u_n}=\frac{u_1+d(n-1)}{u_1u_n}$$
Now replace $d$ by the differences, conveniently:
$$\frac{1}{u_n} + \frac{u_2-u_1}{u_1u_2}+\frac{u_3-u_2}{u_2u_3}+\frac{u_4-u_3}{u_3u_4}+....+\frac{u_n-u_{n-1}}{u_{n-1}u_n}=\frac{u_1+d(n-1)}{u_1u_n}$$
$$\frac{1}{u_n} + \frac{1}{u_1}-\frac{1}{u_2}+\frac{1}{u_2}-\frac{1}{u_3}+-....+\frac{1}{u_{n-1}}-\frac{1}{u_n}=\frac{u_1+d(n-1)}{u_1u_n}$$
This telescopes, giving
$$\frac{1}{u_1}=\frac{u_1+d(n-1)}{u_1u_n}$$
or
$$u_n=u_1+d(n-1)$$
