Question

va rog ex . 11 dau coroana

Answer 1

11. a)
Se calculeaza suma "in sens invers", incepand de la ultima fractie catre prima fractie:
1/(3^100) + 2/(3^100)+ 2/(3^99) +...+ 2/(3^2) + 2/3 = 3/(3^100) + 2/(3^99) +...+ 2/(3^2) + 2/3 = 1/(3^99) + 2/(3^99) +...+ 2/(3^2) + 2/3 = 3/(3^99) + 2/(3^98) +...+ 2/(3^2) + 2/3 = 1/(3^98) + 2/(3^98) +...+ 2/(3^2) + 2/3 = .... = 1/(3^2)+2/(3^2)+2/3 = 3/(3^2)+2/3=1/3+2/3=3/3=1

11. b)
Se calculeaza suma "in sens invers", incepand de la ultima fractie catre prima fractie:
1/(5^100) + 4/(5^100)+ 4/(5^99) +...+ 4/(5^2) + 4/5 = 5/(5^100) + 4/(5^99) +...+ 4/(5^2) + 4/5 = 1/(5^99) + 4/(5^99) +...+ 4/(5^2) + 4/5 = 5/(5^99) + 4/(5^98) +...+ 4/(5^2) + 4/5 = 1/(5^98) + 4/(5^98) +...+ 4/(5^2) + 4/5 = .... = 1/(5^2)+4/(5^2)+4/5 = 5/(5^2)+4/5=1/5+4/5=5/5=1