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!
 A: 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.

A: Your sum $$\sum_{k=4}^{143} \frac{1}{ \sqrt{k} + \sqrt{k+1}} = a - \sqrt{b}$$ converges to $10$ which means that $10 = a - \sqrt b$. After solving this equation we get $a = 10$ $and$ $b = 0$. Hence option(1) is correct. Your approach to rationalize it was correct. If we rationalize it we get $\sqrt(k+1) - \sqrt k$. If we substitute the value of 4 in the equation we get $\sqrt5 - \sqrt4$, we then move on to the next value of $k$ which is $5$ which becomes $\sqrt6 - \sqrt5$. Till now we got, $(\sqrt5 - \sqrt4) + (\sqrt6 - \sqrt5) + (\sqrt7 - \sqrt6) + ... + (\sqrt144 - \sqrt143)$. Now we can cancel out the $\sqrt5$ of the first bracket from the $\sqrt5$ of the second bracket and we will cancel out each and every term. After the cancellation we get $-\sqrt4 + \sqrt144$ and we get 10, which is your answer. Now we can equate it with $a - \sqrt b$ and we get $a = 10$ and $b = 0$
