Question

se dau nr. n1=1+2+3+...+29 n2=10+20+30+...+290 n3=100+200++300+...+2900


sa se afle suma lor adica n1+n2+n3

va rooooog dau coroana​

Answer 1

n₁ = 1+2+3+...+29

n₂ = 10+20+30+...+290

n₃ = 100+200+300+...+2900

n₁ + n₂ + n₃ =

= (1+2+3+...+29) + 10·(1+2+3+...+29) + 100·(1+2+3+...+29)

= (1+2+3+...+29)·(1+10+100)

= (1+2+3+...+29)·111

= (29·30):2 · 111

= 29·15·111

= 435·111

= 48285

Answer 2

Răspuns:

Explicație pas cu pas:

n1=1+2+3+...+29=29×30/2=29×15=435

n2=10+20+30+...+290 =10(1+2+3+.....+29)=10×435

n3=100+200++300+...+2900=100(1+2+3+.....+29)=100×435

n1+n2+n3=435(1+10+100)=435×111=48 285