Odgovor:
The Desna možnost je (c) #2.#
Pojasnilo:
Upoštevajte, da, #AA n v 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 v NN) …… (ast) #.
Uporaba # (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#.
Torej Desna možnost je (c) #2.#
