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: 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 $\sum$.
 A: 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 ;)
A: 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.
