Question

32 aratati ca
a. 0,5<1/1×2+1/2×3+....+1/99×100<1
b. 2/5<2/1×3+2/3×5+2/5×7+....+2/100×101<1
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Answer 1

Explicație pas cu pas:

a)

$\frac{1}{1\cdot 2} + \frac{1}{2 \cdot 3} + ... + \frac{1}{98 \cdot 99} + \frac{1}{99 \cdot 100} \\ = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{98} - \frac{1}{99} + \frac{1}{99} - \frac{1}{100} \\ = \frac{1}{1} - \frac{1}{100} = \frac{100 - 1}{100} = \frac{99}{100}$

$0,5 = \frac{50}{100} < \frac{99}{100} < \frac{100}{100} = 1$

b)

$\frac{2}{1\cdot 3} + \frac{2}{3 \cdot 5} + \frac{2}{5 \cdot 7} + ... + \frac{2}{97 \cdot 99} + \frac{1}{99 \cdot 101} = \\ = \frac{3 - 1}{1\cdot 3} + \frac{5 - 3}{3 \cdot 5} + \frac{7 - 5}{5 \cdot 7} + ... + \frac{99 - 97}{97 \cdot 99} + \frac{101 - 99}{99 \cdot 101} \\ = \frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + ... + \frac{1}{97} - \frac{1}{99} + \frac{1}{99} - \frac{1}{101} \\ = \frac{1}{1} - \frac{1}{101} = \frac{101 - 1}{101} = \frac{100}{101}$

$\frac{2}{5} < \frac{100}{101} < 1$

q.e.d.