<prompt> Hebu tau frac1 sqrt52. Findsum_n 0infty fracfloor taun rceil2n.Kumbuka: Kwa nambari halisi x, floor x rceil inaashiria nambari kamili iliyo karibu zaidi na x.  <response>: Kumbuka kwamba floor tau0 rceil floor 1 rceil 1 na floor tau rceil 2.Acha sigma frac1 - sqrt52, na uruhusu L_n taun sigman. KishaanzaalignL_n taun sigman tau sigmataun - 1 sigman - 1 - tau sigma taun - 2 sigman - 2 L_n - 1 L_n - 2.endalignPia, L_0 2 na L_2 1, kwa hivyo L_n ni nambari kamili kwa n zote ge 0.Zaidi ya hayo,sigma2 frac3 - sqrt52 frac12,hivyo kwa n ge 2, sigman frac12. Kwa hiyo,sakafu taun rceil L_nkwa n zote ge 2.HebuS fracL_222 fracL_323 fracL_424 dotsb.KishaanzaalignS fracL_222 fracL_323 fracL_424 dotsb fracL_0 L_122 fracL_1 L_223 fracL_2 L_324 dotsb kushoto fracL_022 fracL_123 fracL_224 dotsb kulia left frac L_122 fracL_223 fracL_324 dotsb kulia kushoto frac12 frac18 fracS4 kulia kushoto frac14 frac S2 kulia.endalignKutatua, tunapata S frac72.Kwa hiyo,sum_n 0infty fracfloor taun rceil2n 1 frac22 frac72 frac112. 