Question

a=3*(1+3+5+...+99)-5000

Answer 1

1+3+5+...+2n-1=n^2

1+3+5+...+99=?

2n-1=99=>n=(99+1)/2=100/2=50

=>1+3+5+...+99=50^2

1+3+5+...+99=2500

A=3*2500-5000

A=7500-5000

A=2500

Answer 2

Răspuns:

Explicație pas cu pas:

Observam ca in paranteza ai trecute toate numerele impare de la 1 la 100. Insa cum noi stim doar Gauss, ne folosim de urmatorul artificiu: vom rescrie paranteza dupa cum urmeaza

(1+2+3+4+5+6+....+98+99+100-2-4-6-...-98-100) <=> [(1+2+3+4+5+6+...+98+99+100) - (2+4+6+...98+100)] =

(100*101/2) - 2 * (1+2+3+...+49+50)=

50*101 - 2 * 50*51/2 = 50*101-50*51=50*(101-51)=50*50 = 2500

revenim la ecuatia initiala si vom avea:

a= 3*2500-5000=7500-5000=2500