Question

2+4+6+...+2016

1+3+5+...+2017

(22+44+...+2024):(11+22+...+1012)

Answer 1

a.   = 2 · ( 1 + 2+ .. + 1008 ) = 2  · ( 1 + 1008 ) ·1008 /2 = 
        = 1009 · 1008 = 1017072

b.      =(  22 ·1  + 22·2 + .. + 22 · 92 ) : ( 11·1   + 11·2 + .. + 11 · 92) = 
 = 22 · ( 1 + 2 + .. + 92 ) : 11 · ( 1 + 2+ .. + 92) = 
  = 22  :11   · ( 1 + 2 +.. + 92) : ( 1 +2 + .. + 92) = 
   =   2                                  ↓
                                        = 1 
=2 ·1 = 2

Answer 2

$\displaystyle a).2+4+6+...+2016=2(1+2+3+...+1008)= \\ \\ =2 \times \frac{1008(1008+1)}{2} =\not2 \times \frac{1008 \times 1009}{\not2} = 1017072 \\ \\ b).1+3+5+...+2017 = \\ \\ = 1+2+3+4+5+...+2017-(2+4+6+...+2016)= \\ \\ = \frac{2017 \times 2018}{2} -2(1+2+3+...+1008)= \\ \\ =2035153-\not2 \times \frac{1008 \times 1009}{\not2} =2035153-1017072=1018081$

$\displaystyle c).(22+44+...+2024):(11+22+...+1012)= \\ \\ =22(1+2+...+92):11(1+2+...+92)= \\ \\ =22 \times \frac{92(92+1)}{2} :11 \times \frac{92(92+1)}{2} =22 \times \frac{92 \times 93}{2} :11 \times \frac{92 \times 93}{2} = \\ \\ =22 \times \frac{8556}{2} :11 \times \frac{8556}{2} = \frac{188232}{2} : \frac{94116}{2} =94116:47058=2$