Question

Exercitiul 21 punctul b,c,d

Answer 1

a)
nr termeni  
8-2=6
(398-2):6+1=396:6+1=66+1=67
termen de mijloc
(2+398) :2=200
deci vom avea 33 de grupari de genul:
2+398=400
8+392=400
si asa mai departe
2+8+14+20+....+398=33x400+200=13400

c)
nr termeni
2009-10+1=2000
10+11+12+13+....+2009=10+10+1+10+2+10+3+....+10+1999=
=10*2000+(1+2+3+...+1999)=20000+1999*2000/2=
=20000+1999000=2019000

d)
nr termeni
(2009-11):2+1=999+1=1000
putem sa ii grupa astfel   
11+2009=2020
13+2007=2020
si asa mai departe  deci vom avea 100:5=2=500 de astfel de sume
11+13+15+17....+2009=500×20201=1010000