Question

12. Calculaţi: a) 1+2+3+...+9+10; c) 2+4 +6+ ... +98 + 100; b) 1+2+3+...+49 + 50; d) 51 +52 +53 + ... +99 + 100. Rapid
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Answer 1

$a) \: \: 1 + 2 + ... + 10 = \frac{10 \times 11}{2} = \underline{55}$

$b) \: \: 1 +2 + ... + 49 = \frac{49 \times 50}{2} = \underline{1225}$

$c) \: \: 2 + 4 + ... + 98 = 2(1 + 2 + ... + 49) = 2 \times 1225 = \underline{2450}$

$d) \: \: 51 + 52 + ... + 100 = (1 + 2 + ... + 100 )- (1 + 2 + ... + 50) = \frac{100 \times 101}{2} - \frac{50 \times 50}{2} = 50 \times 101 - 25 \times 50 = 5050 - 1250 = \underline{3800}$