The sum $\frac{1}{2\sqrt1+1\sqrt 2}+\frac{1}{3\sqrt2+2\sqrt 3}+\cdots +\frac{1}{100\sqrt{99}+99\sqrt{100}}$ is equal to:

Question

From an exercise of a textbook of an high school of 15 years old, I have this partial sum

$$S_n=\frac{1}{2\sqrt1+1\sqrt 2}+\frac{1}{3\sqrt2+2\sqrt 3}+\cdots +\frac{1}{100\sqrt{99}+99\sqrt{100}}$$

Considering that every fraction is $<1$ I have seen that the general term is $$\frac{1}{100\sqrt{99}+99\sqrt{100}}=\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}$$$$=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n^2+n}=\dfrac{1}{\sqrt{n}} - \dfrac{1}{\sqrt{n+1}}$$

$$A) \frac{999}{1000}, \quad B) \frac{99}{100}, \quad C) \frac{9}{10}, \quad D) 9, \quad E) 1$$

How should I arrive at the result without to use the calculator? My students not have studies the $\sum$.

Answer 1

The general term $$\begin{align} \frac{1}{(n+1)\sqrt{n}+ n\sqrt{n+1}} &= \frac{1}{\sqrt{n(n+1)}}\cdot\frac{1}{\sqrt{n+1}+\sqrt{n}} \\ &=\frac{1}{\sqrt{n(n+1)}}\cdot\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\\ &=\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \end{align}$$

Make the sum, the result must be equal to $$\sum_{n=1}^{99}\frac{1}{(n+1)\sqrt{n}+ n\sqrt{n+1}} =\sum_{n=1}^{99}\left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \right)= \frac{1}{\sqrt{1}}-\frac{1}{\sqrt{100}} = \frac{9}{10}$$

$C$ is the final answer ;)

Answer 2

We can split the terms into two that can be eliminated between adjacent terms. \begin{align} \sum_{i=1}^{99}{\frac{1}{\left( i+1 \right) \sqrt{i}+i\sqrt{i+1}}}&=\sum_{i=1}^{99}{\frac{\left( i+1 \right) \sqrt{i}-i\sqrt{i+1}}{\left( i+1 \right) ^2i-i^2\left( i+1 \right)}}=\sum_{i=1}^{99}{\frac{\left( i+1 \right) \sqrt{i}-i\sqrt{i+1}}{i^2+i}} \\ &=\sum_{i=1}^{99}{\left( \frac{\sqrt{i}}{i}-\frac{\sqrt{i+1}}{i+1} \right)} \\ &=1-\frac{\sqrt{100}}{100}=\frac{9}{10} \end{align}

So the answer is C.

P.S. To write it without the symbol $\sum$, analyzing each term and then writing the equation termwise would help.