Question

fie numerele a=2/9+3/12+...+...+1003/3012,b=3/9+4/12+...1014/3012,c==4/9+5/12+....+1005/3012. calc. O treime din suma numerelor a,b,c
PLEASEEEEE!!!​

Answer 1

Răspuns:

Explicație pas cu pas:

Fiecare nr. a, b, c contine acelasi nr.

de termeni :  

1003-2 +1 = 1002

1004-3 +1 = 1002

1005-4 +1 = 1002

2/9+3/12+4/15....+1003/3012 +  

3/9+4/12+5/15... +1004/3012 +

4/9+5/12+6/15....+1005/3012 =

2/9+3/9+4/9 + 3/12+4/12+5/12 + 4/15+5/15+6/15+...

          ...1003/3012+1004/3012+1005/3012 =

(2+3+4)/9 +(3+4+5)12 + (4+5+6)/15 + ...

          ... (1003+1004+1005)3012 =

1 + 1 + 1 +...+1 = 3*1002 = 3006

1/3*3006 = 1002