Odgovor:
Desna opcija je (C) #2.#
Obrazloženje:
Imajte na umu da, #AA n u NN, 1 / (sqrt (n + 1) + sqrtn) #, # = 1 / (sqrt (n + 1) + sqrtn) xx {(sqrt (n + 1) -sqrtn)} / {(sqrt (n + 1) -sqrtn)} #, # = {(Sqrt (n + 1) -sqrtn)} / {(n + 1) -n} #.
Tako, # 1 / (sqrtn + sqrt (n + 1)) = sqrt (n + 1) -sqrtn; (n u NN) …… (ast) #.
koristeći # (ast) "za" n = 1,2, …, 8 #, imamo, # 1 / (+ sqrt1 sqrt2) + 1 / (+ sqrt2 sqrt3) + 1 / (+ sqrt3 sqrt4) + … + 1 / (+ sqrt8 sqrt9) #, # = (Cancelsqrt2-sqrt1) + (cancelsqrt3-cancelsqrt2) + (cancelsqrt4-cancelsqrt3) + … + (sqrt9-cancelsqrt8) #
# = Sqrt9-sqrt1 #, #=3-1#, #2#.
Dakle, Desna opcija je (C) #2.#