Question

a) 1+2+3+...+25
b125 +26+22+...+ 75
< 1+2+3+. +88
d) 32 +33 +34 + +90
вз 70+в 9+68 ...+31 2+1
+2 100+ 99 +98 +...+52 +54

Answer 1

Explicație pas cu pas:

sume Gauss

$\boxed {1 + 2 + 3 + ... +n = \frac{n(n + 1)}{2}}$

a)

$1 + 2 + 3 + ... + 25 = \frac{25(25 + 1)}{2} = \frac{25 \times 26}{2} = 325$

b)

$25 + 26 + 26 + ... + 75 = (1 + 2 + 3 + ... + 75) - (1 + 2 + 3 + ... + 24) = \frac{75 \times 76}{2} - \frac{24 \times 25}{2} = 2850 - 300 = 2550$

c)

$1 + 2 + 3 + ... + 88 = \frac{88 \times 89}{2} = 3916$

d)

$32 + 33 + 34 + ... + 90 = (1 + 2 + 3 + ...90) - (1 + 2 + 3 + ... + 31) = \frac{90 \times 91}{2} - \frac{31 \times 32}{2} = 4095 - 496 = 3599$