Question

a) 2+4+6+...+2014
b) 1+3+5+...+1001

Answer 1

suma Gauss
se da 2 factor comun:2(1+2+3+4+...+2014)=2[n(n+1)/2
=2*[2014(2014+1)]:2=2*(2014*2015):2=2014*2015=4058210

a) 1+3+5+...+1001=
Aceasta suma nu este una Gauss, trebuie sa observam din cat in cat cresc numerele. In cazul de fata cresc din 2 in 2 scrie numerele ca fiind un produs de 2 * y + 1, unde y
1 = 2 * 0 + 1
3 = 2 * 1 + 1
5 = 2 * 2 + 1
tot asa pina la1001

Answer 2

$\displaystyle a).2+4+6+...+2014=2(1+2+3+...+1007)= \\ \\ =2 \times \frac{1007(1007+1)}{2} =2 \times \frac{1007 \times 1008}{2} =\not2 \times \frac{1015056}{\not2} =1015056$

$\displaystyle b).1+3+5+...+1001= \\ \\ =1+2+3+4+5+...+1001-(2+4+6+...+1000)= \\ \\ = \frac{1001(1001+1)}{2} -2(1+2+3+...+500)= \\ \\ = \frac{1001 \times 1002}{2} -2 \times \frac{500(500+1)}{2} = \frac{1003002}{2} -2 \times \frac{500 \times 501}{2} = \\ \\ =501501-\not 2 \times \frac{250500}{\not2} =501501-250500=251001$