Question

Как правильно решить 30+31+32++38+39+40 11+12+13++87+88+89

Answer 1

$30 + 31 + ... + 39 + 40 = 1 + 2 + ... + 40 - (1 + 2 + ... + 29) =\\\\= \frac{40*41}{2} - \frac{29*30}{2} = 20*41 - 29*15 = \boxed{385}$

$30 + 31 + ... + 39 + 40 =\\\\= 30 + (30 + 1) + ... + (30 + 9) + (30 + 10) =\\\\ = 11*30 + (1 + 2 + ... + 10) = 330 + \frac{10*11}{2} = 330 + 55 = \boxed{385}$

$11+12+ ...+ 89 = 1 + 2 + ... + 89 - ( 1 + 2 + ... + 10) = \\\\ = \frac{89*90}{2} - \frac{11*10}{2} = 89*45 - 11*5 = \boxed{3950}$

Если рассматривать через "призму" арифметических прогрессий, то решить можно так:

$a_1 = 11, \ n = (89 - 11) + 1 = 79, \ a_{79} = 89, \ d = 1\\\\ S_{79} = \frac{a_1 + a_{79}}{2}*79 = \frac{89 + 11}{2}*79 = \frac{100*79}{2} = \frac{7900}{2} = \boxed{3950}$

$a_1 = 30, \ n = (40 - 30) + 1 = 11, \ a_{11} = 40, \ d = 1\\\\ S_{11} = \frac{a_1 + a_{11}}{2}*11 = \frac{30 + 40}{2}*11 = \frac{70*11}{2} =\frac{770}{2} = \boxed{385}$

Answer 2

S=(s1+sn)*n\2
(11+89)*79\2=3950