Need help with an arithmetic sequence proving question It is given that $a_1, a_2, a_3, \ldots ,a_n$ are consecutive terms of an Arithmetic progression. I have to prove that
$$\sum_{k=2}^n (\sqrt{a_{(k-1)}} + \sqrt{a_k} )^{-1} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n} }$$
Using Mathematical induction I showed that it is true for $n = 2$.
Then, assuming that it works for $n = m$, I took the case wherein $n = m+1\ldots$
Using this the left hand side of the equation is :
$$\frac{m-1}{\sqrt{a_1} + \sqrt{a_m} } + (\sqrt{a_{m}} + \sqrt{a_{m+1}} )^{-1} $$
and the right hand side would be 
$$\frac{m}{\sqrt{a_1} + \sqrt{a_{m+1}} }$$
How do I prove that the LHS = RHS??. I tried squaring both numerator and denominator, as well as using the properties of an arithmetic sequence but I havent been able to simplify the algebra.

Actually the induction is not such a mess:
$$
\frac{m-1}{\sqrt{a_1} + \sqrt{a_m} } + (\sqrt{a_{m}} + \sqrt{a_{m+1}} )^{-1}=\frac{(m-1)(\sqrt{a_m} -\sqrt{a_1})}{a_m - a_1 } + \frac{\sqrt{a_{m + 1}} - \sqrt{a_{m}} }{a_{m+1} - a_m}=\frac{(m-1)(\sqrt{a_m} -\sqrt{a_1})}{d(m - 1)}+ \frac{\sqrt{a_{m + 1}} - \sqrt{a_{m}} }{d}=\frac{(\sqrt{a_{m+1}} -\sqrt{a_1})}{d}=\frac{(a_{m+1} -{a_1})}{d(\sqrt{a_{m+1}} +\sqrt{a_1})}=\frac{m}{\sqrt{a_1} + \sqrt{a_{m+1}} }
$$
where d = common difference of the AP.
So LHS = RHS
