Meskipun kompetisi Olimpiade Matematika dilakukan dengan memecahkan
masalah-masalah, sistem Olimpiade Matematika dan kursus pelatihan terkait
tidak hanya melibatkan teknik-teknik pemecahan masalah matematika. Secara
ketat, itu adalah sistem pendidikan matematika yang berfokus pada kemajuan.
Untuk membimbing siswa yang tertarik pada matematika dan memiliki potensi
untuk memasuki dunia matematika Olimpiade, agar kemampuan matematika mereka
dapat ditingkatkan dengan efisien dan komprehensif, penting untuk meningkatkan
pemikiran matematika mereka dan kemampuan teknis dalam memecahkan masalah
matematika.
Seorang siswa yang sangat baik harus mampu berpikir secara fleksibel dan
ketat. Di sini, kemampuan dalam penalaran logika formal merupakan komponen
dasar yang penting. Namun, itu bukanlah komponen utama. Berpikir matematika
juga mencakup aspek-aspek kunci lainnya, seperti dimulai dari intuisi dan
memahami inti subjek melalui prediksi, induksi, imajinasi, konstruksi, desain,
dan kemampuan kreatif mereka. Selain itu, kemampuan untuk mengubah hal-hal
konkret menjadi abstrak dan sebaliknya juga diperlukan.
Kemampuan teknis dalam memecahkan masalah matematika tidak hanya melibatkan
melakukan perhitungan dan pembuktian yang akurat dan terampil dengan metode
standar yang tersedia, tetapi juga teknik-teknik yang tidak lazim dan kreatif.
Jelas bahwa kurikulum biasa dalam pendidikan matematika tidak dapat memenuhi
persyaratan di atas, oleh karena itu buku-buku pelatihan Olimpiade Matematika
harus mandiri secara dasar.
Buku ini didasarkan pada catatan kuliah yang digunakan oleh editor selama 15
tahun terakhir untuk kursus pelatihan Olimpiade di beberapa sekolah di
Singapura, seperti Victoria Junior College, Hwa Chong Institution, Nanyang
Girls High School, dan Dunman High School. Cakupan dan kedalamannya jauh
melebihi kurikulum biasa, dan memperkenalkan banyak konsep dan metode
matematika modern.
Inti dari setiap kuliah adalah konsep, teori, dan metode pemecahan masalah
matematika. Contoh-contoh digunakan untuk menjelaskan dan memperkaya
kuliah-kuliah tersebut, serta menunjukkan aplikasinya. Dan dari situ, beberapa
pertanyaan disertakan untuk pembaca mencoba. Solusi terperinci disediakan
dalam buku ini.
Contoh-contoh yang diberikan tidak terlalu rumit sehingga pembaca dapat lebih
mudah memahaminya. Namun, soal-soal latihan mencakup banyak soal yang diambil
dari Kompetisi-kompetisi tersebut dapat digunakan oleh siswa untuk menguji
diri mereka sendiri. Soal-soal tersebut berasal dari berbagai negara, seperti
China, Rusia, Amerika Serikat, dan Singapura. Terutama, terdapat banyak soal
dari China bagi mereka yang ingin memahami lebih baik Olimpiade Matematika di
sana. Soal-soal tersebut dibagi menjadi dua bagian.
Bagian A digunakan untuk latihan siswa, sementara Bagian B menguji kemampuan
siswa dalam menerapkan pengetahuan mereka dalam memecahkan soal-soal kompetisi
nyata.
Setiap volume dapat digunakan untuk kursus pelatihan selama beberapa minggu
dengan beberapa jam per minggu. Soal-soal ujian tidak dianggap sebagai bagian
dari kuliah, karena siswa dapat menyelesaikannya sendiri.
Lecture Notes on Mathematical Olympiad Courses - World Scientific- Xu Jiagu For Junior Section Vol. 1
Operations on Rational Numbers
1. Basic Rules on Addition, Subtraction, Multiplication, Division
Commutative Law: $a+b=b+a \quad a b=b a$
Associative Law: $a+b+c=a+(b+c) \quad(a b) c=a(b c)$
Distributive Law: $a c+b c=(a+b) c=c(a+b)$
2. Rule for Removing Brackets
For any rational numbers $x, y$,
(i) $x+(y)=x+y, \quad x+(-y)=x-y$;
(ii) $x-(y)=x-y, x-(-y)=x+y$.
(iii) $x \times(-y)=-x y ; \quad(-x) \times y=-x y ; \quad(-x) \times(-y)=x y$; $(-1)^n=-1$ for odd $n,(-1)^n=1$ for even $n$.
(iv) If the denominators of the following expressions are all not zeros, then
$$
\frac{x}{-y}=-\frac{x}{y} ; \quad \frac{-x}{y}=-\frac{x}{y} ; \quad \frac{-x}{-y}=\frac{x}{y} .
$$
Associative Law: $a+b+c=a+(b+c) \quad(a b) c=a(b c)$
Distributive Law: $a c+b c=(a+b) c=c(a+b)$
2. Rule for Removing Brackets
For any rational numbers $x, y$,
(i) $x+(y)=x+y, \quad x+(-y)=x-y$;
(ii) $x-(y)=x-y, x-(-y)=x+y$.
(iii) $x \times(-y)=-x y ; \quad(-x) \times y=-x y ; \quad(-x) \times(-y)=x y$; $(-1)^n=-1$ for odd $n,(-1)^n=1$ for even $n$.
(iv) If the denominators of the following expressions are all not zeros, then
$$
\frac{x}{-y}=-\frac{x}{y} ; \quad \frac{-x}{y}=-\frac{x}{y} ; \quad \frac{-x}{-y}=\frac{x}{y} .
$$
3. Ingenious Ways for Calculating
- Make a telescopic sum by using the following expressions:
$$
\begin{aligned}
& \frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}, \\
& \frac{1}{k(k+m)}=\frac{1}{m}\left(\frac{1}{k}-\frac{1}{k+m}\right), \\
& \frac{1}{k(k+1)(k+2)}=\frac{1}{2}\left[\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}\right] .
\end{aligned}
$$
- By use of the following formulae:
$$
\begin{aligned}
& (a \pm b)^2=a^2+2 a b+b^2 \\
& a^2-b^2=(a-b)(a+b) \\
& a^3+b^3=(a+b)\left(a^2-a b+b^2\right) \\
& a^3-b^3=(a-b)\left(a^2+a b+b^2\right), \text { etc. }
\end{aligned}
$$
$\boxed{Examples}$
Example 1.
Evaluate $(-5)^2 \times\left(-\frac{1}{5}\right)^3-2^3
\div\left(-\frac{1}{2}\right)^2-(-1)^{1999}$
Solution
$$
\begin{aligned}
&(-5)^2 \times\left(-\frac{1}{5}\right)^3-2^3 \div\left(-\frac{1}{2}\right)^2-(-1)^{1999} \\
&=5^2 \times\left(-\frac{1}{125}\right)-8 \div \frac{1}{4}-(-1) \\
&=-\frac{1}{5}-8 \times 4+1=-\frac{1}{5}-31=-31 \frac{1}{5}
\end{aligned}
$$
Example 2.
$$
\begin{aligned}
&(-5)^2 \times\left(-\frac{1}{5}\right)^3-2^3 \div\left(-\frac{1}{2}\right)^2-(-1)^{1999} \\
&=5^2 \times\left(-\frac{1}{125}\right)-8 \div \frac{1}{4}-(-1) \\
&=-\frac{1}{5}-8 \times 4+1=-\frac{1}{5}-31=-31 \frac{1}{5}
\end{aligned}
$$
Example 2.
There are five operational expressions below:
(i) $(2 \times 3 \times 5 \times 7)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}\right)$;
(ii) $(-0.125)^7 \cdot 8^8$;
(iii) $(-11)+(-33)-(-55)-(-66)-(-77)-(-88)$;
(iv) $\left(-\frac{75}{13}\right)^2+\left(\frac{37}{13}\right)^2$;
(v) $\left[\left(-\frac{6}{7}\right)^7+\left(-\frac{4}{5}\right) \times\left(-\frac{4}{9}\right) \times \frac{16}{81}\right] \times\left(9 \frac{246}{247}-0.666\right)$.
Then the expression with maximal value is
(A) (i),
(B) (iii),
(C) (iv),
(D) (v).
Solution
(i) $(2 \times 3 \times 5 \times 7)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}\right)$
$=105+70+42+30=247$;
(ii) $(-0.125)^7 \cdot 8^8=-(0.125 \times 8)^7 \times 8=-8$;
(iii) $(-11)+(-33)-(-55)-(-66)-(-77)-(-88)$
$$
=-11-33+55+66+77+88=11 \times 22=242 \text {; }
$$
(iv) $\left(-\frac{75}{13}\right)^2+\left(\frac{37}{13}\right)^2<6^2+3^2=45$;
(v)
$$
\begin{aligned}
& {\left[\left(-\frac{6}{7}\right)^7+\left(-\frac{4}{5}\right) \times\left(-\frac{4}{9}\right) \times \frac{16}{81}\right] \times\left(9 \frac{246}{247}-0.666\right)} \\
& <1 \times 10=10
\end{aligned}
$$
Thus, the answer is (A).
Example 3.
(i) $(2 \times 3 \times 5 \times 7)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}\right)$;
(ii) $(-0.125)^7 \cdot 8^8$;
(iii) $(-11)+(-33)-(-55)-(-66)-(-77)-(-88)$;
(iv) $\left(-\frac{75}{13}\right)^2+\left(\frac{37}{13}\right)^2$;
(v) $\left[\left(-\frac{6}{7}\right)^7+\left(-\frac{4}{5}\right) \times\left(-\frac{4}{9}\right) \times \frac{16}{81}\right] \times\left(9 \frac{246}{247}-0.666\right)$.
Then the expression with maximal value is
(A) (i),
(B) (iii),
(C) (iv),
(D) (v).
Solution
(i) $(2 \times 3 \times 5 \times 7)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}\right)$
$=105+70+42+30=247$;
(ii) $(-0.125)^7 \cdot 8^8=-(0.125 \times 8)^7 \times 8=-8$;
(iii) $(-11)+(-33)-(-55)-(-66)-(-77)-(-88)$
$$
=-11-33+55+66+77+88=11 \times 22=242 \text {; }
$$
(iv) $\left(-\frac{75}{13}\right)^2+\left(\frac{37}{13}\right)^2<6^2+3^2=45$;
(v)
$$
\begin{aligned}
& {\left[\left(-\frac{6}{7}\right)^7+\left(-\frac{4}{5}\right) \times\left(-\frac{4}{9}\right) \times \frac{16}{81}\right] \times\left(9 \frac{246}{247}-0.666\right)} \\
& <1 \times 10=10
\end{aligned}
$$
Thus, the answer is (A).
Example 3.
$123456789 \times 999999999=$
Solution
$$
\begin{aligned}
& 123456789 \times 999999999=123456789 \times(1000000000-1) \\
& =123456789000000000-123456789=123456788876543211 .
\end{aligned}
$$
Example 4.
Solution
$$
\begin{aligned}
& 123456789 \times 999999999=123456789 \times(1000000000-1) \\
& =123456789000000000-123456789=123456788876543211 .
\end{aligned}
$$
Example 4.
The value of $\dfrac{13579}{(-13579)^2+(-13578)(13580)}$ is
(A) 1 ,
(B) 13579 ,
(C) -1 .
(D) -13578 .
Solution By use of $(a-b)(a+b)=a^2-b^2$, we have
$$
\begin{aligned}
& \frac{13579}{(-13579)^2+(-13578)(13580)} \\
& =\frac{13579}{(13579)^2-\left(13579^2-1\right)}=13579 .
\end{aligned}
$$
The answer is (B).
(A) 1 ,
(B) 13579 ,
(C) -1 .
(D) -13578 .
Solution By use of $(a-b)(a+b)=a^2-b^2$, we have
$$
\begin{aligned}
& \frac{13579}{(-13579)^2+(-13578)(13580)} \\
& =\frac{13579}{(13579)^2-\left(13579^2-1\right)}=13579 .
\end{aligned}
$$
The answer is (B).
Example 5.
$\dfrac{83^2+17^3}{83 \times 66+17^2}=$
Solution By use of the formula $a^3+b^3=(a+b)\left(a^2-a b+b^2\right)$,
$$
\begin{aligned}
\dfrac{83^2+17^3}{83 \times 66+17^2} & =\dfrac{(83+17)\left(83^2-83 \times 17+17^2\right)}{83 \times 66+17^2} \\
& =\dfrac{100 \times\left(83 \times 66+17^2\right)}{83 \times 66+17^2}=100 .
\end{aligned}
$$
Solution By use of the formula $a^3+b^3=(a+b)\left(a^2-a b+b^2\right)$,
$$
\begin{aligned}
\dfrac{83^2+17^3}{83 \times 66+17^2} & =\dfrac{(83+17)\left(83^2-83 \times 17+17^2\right)}{83 \times 66+17^2} \\
& =\dfrac{100 \times\left(83 \times 66+17^2\right)}{83 \times 66+17^2}=100 .
\end{aligned}
$$
Example 6.
Evaluate
$$
\dfrac{(4 \times 7+2)(6 \times 9+2)(8 \times 11+2) \cdots(100 \times 103+2)}{(5 \times 8+2)(7 \times 10+2)(9 \times 12+2) \cdots \cdots(99 \times 102+2)}
$$
Solution From $n(n+3)+2=n^2+3 n+2=(n+1)(n+2)$ for any integer $n$, we have
$$
\begin{aligned}
& \frac{(4 \times 7+2)(6 \times 9+2)(8 \times 11+2) \cdots \cdots(100 \times 103+2)}{(5 \times 8+2)(7 \times 10+2)(9 \times 12+2) \cdots \cdots(99 \times 102+2)} \\
& =\frac{(5 \times 6)(7 \times 8)(9 \times 10) \cdots(101 \times 102)}{(6 \times 7)(8 \times 9)(10 \times 11) \cdots(100 \times 101)} \\
& =5 \times 102=510 .
\end{aligned}
$$
Example 7.
$$
\dfrac{(4 \times 7+2)(6 \times 9+2)(8 \times 11+2) \cdots(100 \times 103+2)}{(5 \times 8+2)(7 \times 10+2)(9 \times 12+2) \cdots \cdots(99 \times 102+2)}
$$
Solution From $n(n+3)+2=n^2+3 n+2=(n+1)(n+2)$ for any integer $n$, we have
$$
\begin{aligned}
& \frac{(4 \times 7+2)(6 \times 9+2)(8 \times 11+2) \cdots \cdots(100 \times 103+2)}{(5 \times 8+2)(7 \times 10+2)(9 \times 12+2) \cdots \cdots(99 \times 102+2)} \\
& =\frac{(5 \times 6)(7 \times 8)(9 \times 10) \cdots(101 \times 102)}{(6 \times 7)(8 \times 9)(10 \times 11) \cdots(100 \times 101)} \\
& =5 \times 102=510 .
\end{aligned}
$$
Example 7.
$\frac{20092008^2}{20092007^2+20092009^2-2}=$
$$
\begin{aligned}
& \text { Solution } \frac{20092008^2}{20092007^2+20092009^2-2} \\
& \quad=\frac{20092008^2}{\left(20092007^2-1\right)+\left(20092009^2-1\right)} \\
& \quad=\frac{20092008^2}{(20092006)(20092008)+(20092008)(20092010)} \\
& \quad=\frac{20092008^2}{(20092008)(20092006+20092010)}=\frac{20092008^2}{2\left(20092008^2\right)}=\frac{1}{2} .
\end{aligned}
$$
$$
\begin{aligned}
& \text { Solution } \frac{20092008^2}{20092007^2+20092009^2-2} \\
& \quad=\frac{20092008^2}{\left(20092007^2-1\right)+\left(20092009^2-1\right)} \\
& \quad=\frac{20092008^2}{(20092006)(20092008)+(20092008)(20092010)} \\
& \quad=\frac{20092008^2}{(20092008)(20092006+20092010)}=\frac{20092008^2}{2\left(20092008^2\right)}=\frac{1}{2} .
\end{aligned}
$$
Example 8.
$3-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}=$
Solution
$$
\begin{aligned}
& 3-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right) \\
& =3-\left(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\cdots+\frac{1}{7 \times 8}\right) \\
& =3-\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{7}-\frac{1}{8}\right)\right] \\
& =3-\left(1-\frac{1}{8}\right)=2 \frac{1}{8} \text {. } \\
&
\end{aligned}
$$
Solution
$$
\begin{aligned}
& 3-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right) \\
& =3-\left(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\cdots+\frac{1}{7 \times 8}\right) \\
& =3-\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{7}-\frac{1}{8}\right)\right] \\
& =3-\left(1-\frac{1}{8}\right)=2 \frac{1}{8} \text {. } \\
&
\end{aligned}
$$
Example 9.
Evaluate
$\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}$.
Solution Since $\frac{1}{k(k+2)}=\frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+2}\right)$ for any positive integer $k$, so
$$
\begin{aligned}
& \frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143} \\
& =\frac{1}{1 \times 3}+\frac{1}{3 \times 5}+\frac{1}{5 \times 7}+\frac{1}{7 \times 9}+\frac{1}{9 \times 11}+\frac{1}{11 \times 13} \\
& =\frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots+\left(\frac{1}{11}-\frac{1}{13}\right)\right] \\
& =\frac{1}{2} \times\left[1-\frac{1}{13}\right]=\frac{6}{13} .
\end{aligned}
$$
Solution Since $\frac{1}{k(k+2)}=\frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+2}\right)$ for any positive integer $k$, so
$$
\begin{aligned}
& \frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143} \\
& =\frac{1}{1 \times 3}+\frac{1}{3 \times 5}+\frac{1}{5 \times 7}+\frac{1}{7 \times 9}+\frac{1}{9 \times 11}+\frac{1}{11 \times 13} \\
& =\frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots+\left(\frac{1}{11}-\frac{1}{13}\right)\right] \\
& =\frac{1}{2} \times\left[1-\frac{1}{13}\right]=\frac{6}{13} .
\end{aligned}
$$
Example 10.
If $a b<0$, then the relation in sizes of
$(a-b)^2$ and $(a+b)^2$ is
(A) $(a-b)^2<(a+b)^2$
(B) $(a-b)^2=(a+b)^2$
(C) $(a-b)^2>(a+b)^2$,
(D) not determined.
Solution From $(a-b)^2=a^2-2 a b+b^2=a^2+2 a b+b^2-4 a b=$ $(a+b)^2-4 a b>(a+b)^2$, the answer is $(C)$.
(A) $(a-b)^2<(a+b)^2$
(B) $(a-b)^2=(a+b)^2$
(C) $(a-b)^2>(a+b)^2$,
(D) not determined.
Solution From $(a-b)^2=a^2-2 a b+b^2=a^2+2 a b+b^2-4 a b=$ $(a+b)^2-4 a b>(a+b)^2$, the answer is $(C)$.
Example 11.
If $-1<a<0$, then the relation in sizes of
$a^3,-a^3, a^4,-a^4, \frac{1}{a}$, $-\frac{1}{a}$ is
(A) $\frac{1}{a}<-a^4<a^3<-a^3<a^4<-\frac{1}{a}$;
(B) $a<\frac{1}{a}<-a^4<a^4<-\frac{1}{a}<-a^3$,
(C) $\frac{1}{a}<a^3<-a^4<a^4<-a^3<-\frac{1}{a}$;
(D) $\frac{1}{a}<a^3<a^4<-a^4<-a^3<-\frac{1}{a}$.
Solution From $-1<a<0$ we have $0<a^4<-a^3<1<-\frac{1}{a}$, so $-a^4>a^3$ and $-\frac{1}{a}>-a^3$ and $a^4>-a^4$, the answer is $(\mathrm{C})$.
(A) $\frac{1}{a}<-a^4<a^3<-a^3<a^4<-\frac{1}{a}$;
(B) $a<\frac{1}{a}<-a^4<a^4<-\frac{1}{a}<-a^3$,
(C) $\frac{1}{a}<a^3<-a^4<a^4<-a^3<-\frac{1}{a}$;
(D) $\frac{1}{a}<a^3<a^4<-a^4<-a^3<-\frac{1}{a}$.
Solution From $-1<a<0$ we have $0<a^4<-a^3<1<-\frac{1}{a}$, so $-a^4>a^3$ and $-\frac{1}{a}>-a^3$ and $a^4>-a^4$, the answer is $(\mathrm{C})$.
Testing Questions (A)
2. Evaluate $2008 \times 20092009-2009 \times 20082008$.
3. From 2009 subtract half of it at first, then subtract $\frac{1}{3}$ of the remaining number, next subtract $\frac{1}{4}$ of the remaining number, and so on, until $\frac{1}{2009}$ of the remaining number is subtracted. What is the final remaining number?
4. Find the sum $\frac{1}{5 \times 7}+\frac{1}{7 \times 9}+\frac{1}{9 \times 11}+\frac{1}{11 \times 13}+\frac{1}{13 \times 15}$.
5. Find the sum $\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}$.
6. Evaluate
$$
\begin{aligned}
& \left(\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2009}\right)\left(1+\frac{1}{2}+\cdots+\frac{1}{2008}\right) \\
& -\left(1+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2009}\right)\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2008}\right)
\end{aligned}
$$
7. Find the sum $\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots+\frac{1}{1+2+\cdots+51}$.
8. Let $n$ be a positive integer, find the value of
$$
1+\frac{1}{2}+\frac{2}{2}+\frac{1}{2}+\frac{1}{3}+\frac{2}{3}+\frac{3}{3}+\frac{2}{3}+\frac{1}{3}+\cdots+\frac{1}{n}+\frac{2}{n}+\cdots+\frac{n}{n}+\frac{n-1}{n}+\cdots+\frac{1}{n}
$$
9. Evaluate $1^2-2^2+3^2-4^2+\cdots-2008^2+2009^2$.
10. Find the sum $11+192+1993+19994+199995+1999996+19999997+$ $199999998+1999999999$.
Testing Questions (B)
2. After simplification, the value of
$$
\begin{aligned}
& 1-\dfrac{2}{1 \cdot(1+2)}-\dfrac{3}{(1+2)(1+2+3)}-\dfrac{4}{(1+2+3)(1+2+3+4)} \\
& -\cdots-\dfrac{100}{(1+2+\cdots+99)(1+2+\cdots+100)}
\end{aligned}
$$
is a proper fraction in its lowest form. Find the difference of its denominator and numerator.
3. Evaluate $\frac{1}{1 \times 2 \times 3}+\frac{1}{2 \times 3 \times 4}+\cdots+\frac{1}{100 \times 101 \times 102}$.
4. Find the sum
$$
\dfrac{1}{1+1^2+1^4}+\dfrac{2}{1+2^2+2^4}+\dfrac{3}{1+3^2+3^4}+\cdots+\dfrac{50}{1+50^2+50^4}
$$
5. Evaluate the expression
$$
\dfrac{1^2}{1^2-10+50}+\dfrac{2^2}{2^2-20+50}+\cdots+\dfrac{80^2}{80^2-80+50}
$$
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