Prove $\frac{1}{u_{0} u_{1}}+\frac{1}{u_{1} u_{2}}+\ldots+\frac{1}{u_{i} u_{i+1}}+\ldots \ldots+\frac{1}{u_{n} u_{n+1}}=\frac{n+1}{u_{0} u_{n+1}}$ [closed]

Let $\mathrm{u}_{\mathrm{n}}$ be an arithmetic progression (sequence) with common difference $\mathrm{d}\left(\mathrm{u}_{\mathrm{i}+1}=\mathrm{u}_{\mathrm{i}}+\mathrm{d}, \mathrm{i} \geq 0\right)$. Demonstrate that : $$ \frac{1}{u_{0} u_{1}}+\frac{1}{u_{1} u_{2}}+\ldots+\frac{1}{u_{i} u_{i+1}}+\ldots \ldots+\frac{1}{u_{n} u_{n+1}}=\frac{n+1}{u_{0} u_{n+1}} $$

I've tried so hard but I didn't get a result. Any thoughts?

Solution 1:

In case you were wondering how kmitov's comment came about, note that $d=u_1-u_0=u_2-u_1=...=u_{n}-u_{n-1}$

LHS $=\frac{1}{u_1-u_0}\frac{u_1-u_0}{u_1u_0}+\frac{1}{u_2-u_1}\frac{u_2-u_1}{u_2u_1}+...+\frac{1}{u_{n+1}-u_n}\frac{u_{n+1}-u_n}{u_{n+1}u_n}$

$=\frac{1}{d}\left(\frac{1}{u_0}-\frac{1}{u_1}\right)+\frac{1}{d}\left(\frac{1}{u_1}-\frac{1}{u_2}\right)+\frac{1}{d}\left(\frac{1}{u_3}-\frac{1}{u_2}\right)+...+\frac{1}{d}\left(\frac{1}{u_n}-\frac{1}{u_{n+1}}\right)$

Factoring out the $\frac{1}{d}$, the reciprocals cancel out to give LHS = $\frac{1}{d}\left(\frac{1}{u_0}-\frac{1}{u_{n+1}}\right)$

$=\frac{1}{d}\left(\frac{u_{n+1}-u_0}{u_{n+1}u_0}\right)$

We know that $S_n=\frac{n}{2}(2a+(n-1)d)=\frac{n}{2}(a+l)$ where $a$ and $l$ are the first and last terms respectively and $n$ is the number of terms.

Therefore $2a+(n-1)d=a+l$, which means $l-a=(n-1)d$. Note the the number of terms in your series is $n+2$.

$\therefore u_{n+1}-u_0=(n+1)d$

$\therefore$ LHS $=\frac{1}{d}\left(\frac{(n+1)d}{u_{n+1}u_0}\right)$, which then yields your result.