Usage of Mathematical Induction

Question

How do I prove this with Mathematical Induction? Whereby $$u_1, u_2...u_n$$ are all positive and are in an arithmetic progression for $$n\geq2$$

$$\sum\limits_{k=2}^{n}\frac{1}{(u_{k-1})\cdot(u_k)}=\frac{n-1}{(u_1)\cdot(u_n)}$$

Answer 1

Hint

It looks like the results comes from telescoping series

So write

$u_k = u_{k-1} + d$, then $$\frac 1{u_k u_{k-1}} = \frac 1{u_{k-1}^2 + du_{k-1}} = \frac 1d \left(\frac{1}{u_{k-1}} - \frac 1{u_{k-1}+d}\right) = \frac 1d \left(\frac{1}{u_{k-1}} - \frac 1{u_k}\right)$$

Answer 2

If $\sum\limits_{k=2}^m\dfrac1{(u_{k-1})\cdot(u_k)}=\dfrac{m-1}{(u_1)\cdot(u_m)},$

$$\sum\limits_{k=2}^{m+1}\frac1{(u_{k-1})\cdot(u_k)}=\sum\limits_{k=2}^m\frac1{(u_{k-1})\cdot(u_k)}+\frac1{u_mu_{m+1}}=\frac{m-1}{u_1 u_m}+\frac1{u_mu_{m+1}}$$

We need to have $$\frac{m-1}{u_1 u_m}+\frac1{u_mu_{m+1}}=\frac m{u_1u_{m+1}}$$

$$\iff(m-1)u_{m+1}+u_1=mu_m\iff m(u_{m+1} - u_m)=u_{m+1}-u_1$$

Write $u_r=u_0+(r-1)d$ where $d$ is the common difference

Answer 3

Hint $\ $ Just $ $ like $\ \sum\, \frac{1}{(k-1)k}\ $ the sum telescopes

$\qquad \dfrac{1}{k-1} - \dfrac{1}{k}\ =\, \dfrac{\color{#c00}1}{(k\!-\!1)k},\quad \color{#c00}1 = k-(k\!-\!1)$

$\qquad \dfrac{1}{u_{k-1}} - \dfrac{1}{u_k}\ =\, \dfrac{\color{#c00}\delta}{u_{k-1}u_k},\quad \color{#c00}\delta = u_{k}-u_{k-1}\,$ constant for all $\,k,\,$ by $\,u_k$ an AP