Question

Se con nr a=1+2+3+....+2017 ; b= 1²+2²+3²+.......2017²si c= 1×2+2×3+3×4+2017×2018
A)demonstrati ca nr a se divide cu 2017 .
B) calculati a+b-c

Va rog dau coroana

Answer 1

a) aplicam formula pentru suma primilor n termeni de numere naturale consecutive adica n(n+1)/2
     deci 1+2 +....+2017 = 2017*2018/2=  2017*1014  deci suma noastra se divide cu 2017.

b) a+b-c=(1+2+...+2017)+(1²+2²+...2017²)-(1×2+2×3+...+2017×2018)=
=(grupam din primele 2 paranteze termenii in felul urmator(1 cu 1²; 2 cu 2² etc)= 1+1²+2+2²+...+2017+2017² - (1×2+2×3+...+2017×2018) = 
= (dam factor comun 2 cate 2 in felul urmator)=
=1(1+1) + 2(1+2) + 3(1+3)+....+2017(1+2017) - (1×2+2×3+...+2017×2018)=
=(1×2 +2×3 +3×4 + ...+ 2017×2018)-(1×2+2×3+...+2017×2018)=
=(observam ca parantezele sunt egale deci a+b-c=0) = 0