Đáp án:
$a) \frac{E}{F} = 101$
$b) F - 101G = - \frac{100}{101}$
Giải thích các bước giải:
$ a) E + 100 = \frac{100² + 1²}{100.1} + 2 + \frac{99² + 2²}{99.2} + 2 + \frac{98² + 3²}{98.2} + 2 + ...+ \frac{52² + 49²}{52.49} + 2 + \frac{51² + 50²}{51.50} + 2 = \frac{100² + 2.100.1 + 1²}{100.1} + \frac{99² + 2.99.2 + 2²}{99.2} + \frac{98² + 2.98.3+ 3²}{98.3} + ...+ \frac{52² + 2.52.49 + 49²}{52.49} + \frac{51² + 2.51.50+ 50²}{51.50}$
$= \frac{(100 + 1)²}{100.1} + \frac{(99 + 2)²}{99.2} + \frac{(98 + 3)²}{98.3} + ...+ \frac{(52 + 49)²}{52.49} + \frac{(51 + 50)²}{51.50} = 101(\frac{100 + 1}{100.1} + \frac{99 + 2}{99.2} + \frac{98 + 3}{98.3} + ...+ \frac{52 + 49}{52.49} + \frac{51 + 50}{51.50}) = 101(\frac{1}{1} + \frac{1}{100} + \frac{1}{2} + \frac{1}{99} + \frac{1}{3} +\frac{1}{98} + ...+ \frac{1}{49} + \frac{1}{52} + \frac{1}{50} + \frac{1}{51}) = 101[(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} +..+ \frac{1}{100} + \frac{1}{101}) + 1 - \frac{1}{101}] = 101(F + 1 - \frac{1}{101}) = 101F + 100 (1)$
$ ⇒ E = 101F ⇒ \frac{E}{F} = 101$
$ b) E + 100 = \frac{100² + 1²}{100.1} + 2 + \frac{99² + 2²}{99.2} + 2 + \frac{98² + 3²}{98.2} + 2 + ...+ \frac{52² + 49²}{52.49} + 2 + \frac{51² + 50²}{51.50} + 2 = \frac{100² + 2.100.1 + 1²}{100.1} + \frac{99² + 2.99.2 + 2²}{99.2} + \frac{98² + 2.98.3+ 3²}{98.3} + ...+ \frac{52² + 2.52.51 + 49²}{52.49} + \frac{51² + 2.51.50+ 50²}{51.50}$
$= \frac{(100 + 1)²}{100.1} + \frac{(99 + 2)²}{99.2} + \frac{(98 + 3)²}{98.3} + ...+ \frac{(52 + 49)²}{52.49} + \frac{(51 + 50)²}{51.50}$
$= 101²(\frac{1}{100.1} + \frac{1}{99.2} + \frac{1}{98.3} + ...+ \frac{1}{52.49} + \frac{1}{51.50}) = 101²G (2)$
Từ $(1) ; (2) ⇒ 101F + 100 = 101²G ⇒ F - 101G = - \frac{100}{101}$
