If $\sum_{k=4}^{143} \frac{1}{ \sqrt{k} + \sqrt{k+1}} = a - \sqrt{b}$, then $a$ and $b$ respectively are If $\sum_{k=4}^{143} \frac{1}{ \sqrt{k} + \sqrt{k+1}} = a - \sqrt{b}$, then $a$ and $b$ respectively are

*

*10 and 0

*-10 and 4

*10 and 4

*-10 and 0


This question is from the book, Mathematics, Class 9 (The IIT Foundation Series) , page number 1.25. The answer key present in the book says the first option as the correct answer. I need an explanation to solve this question. Till now I have tried to rationalize $\frac{1}{ \sqrt{k} + \sqrt{k+1}}$ found that $\frac{1}{ \sqrt{k} + \sqrt{k+1}} = \sqrt{k+1}-\sqrt{k}$. I am not sure what to do next.
Thanks!

The four alternatives result when calculating $a-\sqrt{b}$ in
\begin{align*}
&(10,0)&10-\sqrt{0}&=10\tag{1.}\\
&(-10,4)&-10-\sqrt{4}&=-12\tag{2.}\\
&(10,4)&10-\sqrt{4}&=8\tag{3.}\\
&(-10,0)&-10-\sqrt{0}&=-10\tag{4.}
\end{align*}

We can now use OPs telescoping approach and obtain
\begin{align*}
\color{blue}{\sum_{k=4}^{143} \frac{1}{ \sqrt{k+1} + \sqrt{k}}}
&=\sum_{k=4}^{143}\left(\sqrt{k+1}-\sqrt{k}\right)\\
&=\sqrt{143+1}-\sqrt{4}\\
&=12-2\\
&\,\,\color{blue}{=10}
\end{align*}
showing (1.) is correct.

