Question

sa se calculexe sumele    
s1 = 1+2+3+...=100
s2=1+2+4+8...=128

Answer 1

1.  1+2+3+4+..........+100 =100×101/2 = 5050

daca, scrie 1+2+3+4+ .............=100  ⇒n(n+1)/2 =100  n(n+1) =200         n∉N  (nu exista 2 numere consecutive al caror produs =200)

2. S2 =1+2+2²+2³+2^4+........2^7

    2S2 =2+2²+2³+................+2^7+2^8

2S2 - S2 = S2 = 2^8 - 1   ⇒  S2 =(2^8 - 1) /2 = 1023/2