Find $x$ for $\left(\frac1{1\times101} + \frac1{2\times102} + \dots +\frac1{10\times110}\right)x = \frac1{1\times11} + \frac1{2\times12}...$ $$\left(\frac1{1\times101} + \frac1{2\times102} + \dots +\frac1{10\times110}\right)x = \frac1{1\times11} + \frac1{2\times12} + \dots +\frac1{100\times110}$$
Find x
My younger sister in grade 5 had this question in a test. But I, a college student, still can't solve this. What a shame :<
 A: One can group the $100$ summands on the right side by the ending digit of the denominators. For example, taking the numbers ending in 2:
$$\frac{1}{2\times 12}+\frac{1}{12 \times 22}+ ... +\frac{1}{92 \times 102}=\\
= \frac{1}{10}\left(\frac{1}{2} -\frac{1}{12}\right) +\frac{1}{10}\left(\frac{1}{12} -\frac{1}{22}\right) +\cdots  +\frac{1}{10}\left(\frac{1}{92}- \frac{1}{ 102}\right) =\\ =\frac{1}{10}\left(\frac{1}{2}-\frac{1}{102}\right)=\\ =10\left(\frac{1}{2 \times 102}\right)\\
$$
Therefore, $x=10$
This can be formalized, of course. But I don't think that this (even without the formalization) is appropiate for a 10 years old...
A: Purely computationally:
$$\sum_1^{10} \frac{1}{k(k+100)} = \frac{59810902182173}{2110393648079145}, $$
and
$$\sum_1^{100} \frac{1}{k(k+10)} = \frac{119621804364346}{422078729615829}. $$
The second numerator is exactly double the first, and the second denominator is exactly a fifth of the first, leading to the surprising (for me) result that
$$x=10.$$
A: $$\frac{1}{k(k+10)}=\frac{a}{k}+\frac{b}{k+10}$$
$$\begin{array}{l}
\frac{a}{k}+\frac{b}{k+10}=\frac{k(a+b)+10a}{k(k+10)}
\end{array}$$
$$k\leftarrow 0, a=\frac{1}{10}$$
$$k\leftarrow 1, b=-\frac{1}{10}$$
$$\boxed{\cfrac{1}{k(k+10)}=\cfrac{1}{10}\left(\cfrac{1}{k}-\cfrac{1}{k+10}\right)}$$

$$\begin{array}{l}
\sum_{k=1}^{100} \frac{1}{k(k+10)}&=\frac{1}{10}\sum_{k=1}^{100} \frac{1}{k}-\cfrac{1}{k+10}\\
&=\frac{1}{10}\left(\sum_{k=1}^{100} \frac{1}{k}-\sum_{k=1}^{100}\cfrac{1}{k+10}\right)\\
&=\frac{1}{10}\left(\sum_{k=1}^{100} \frac{1}{k}-\sum_{k=11}^{110}\cfrac{1}{k}\right)\\
&=\frac{1}{10}\left(\sum_{k=1}^{10} \frac{1}{k}-\sum_{k=101}^{110}\cfrac{1}{k}\right)\\
&=\frac{1}{10}\left(\sum_{k=1}^{10} \frac{1}{k}-\sum_{k=1}^{10}\cfrac{1}{k+100}\right)\\
&=\frac{1}{10}\left(\sum_{k=1}^{10} \frac{1}{k}-\cfrac{1}{k+100}\right)\\
&=\frac{1}{10}\left(\sum_{k=1}^{10} \frac{100}{k(k+100)}\right)\\
\sum_{k=1}^{100} \frac{1}{k(k+10)}&=10\sum_{k=1}^{10} \frac{1}{k(k+100)}
\end{array}$$
