<s>[INST] 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!
 [/INST] 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 $b = 0$
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