Prove that, $\sum_{k=1}^{n-1} \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}.$

Question

The sequence $(a_k)_{k \geqslant 1}$ is an AP. Prove that, for every $n \in \mathbb{Z^+}$, $$\sum_{k=1}^{n-1} \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}.$$

I rationalized it to $$\sum_{k=1}^{n-1}\frac{\sqrt{a_k} - \sqrt{a_{k-1}}}{a_k - a_{k-1}}$$

How do i find the the partial fractions to telescope this ?

Answer 1

Note that the sequence is an $AP$, hence the denominator is a constant.

Hence is is equal to $$\frac1d (\sqrt{a_n} - \sqrt{a_1})=\frac1d\frac{a_n-a_1}{\sqrt{a_1}+\sqrt{a_n}}=\frac{n-1}{\sqrt{a_1}+\sqrt{a_n}}$$

Remark: be careful with the indices. $a_0$ didn't appear in the original sum. It should be kept as $a_{k+1}$ and $a_k$.