<prompt> Acha a_0, a_1, a_2, dots iwe mfuatano usio na kikomo wa nambari halisi kama vile a_0 frac513 na a_n 2 a_n-12 - 1kwa kila nambari kamili n. Acha c iwe nambari ndogo zaidi kwamba kwa kila nambari kamili n, bidhaa ya masharti ya n ya kwanza inakidhi ukosefu wa usawa.a_0 a_1 dotsm a_n - 1 le fracc2n.Ni nini thamani ya 100c, iliyozungushwa hadi nambari kamili iliyo karibu zaidi?  <response>: Bainisha mlolongo theta_n by theta_0 arccos frac513 natheta_n 2 theta_n - 1.Kisha cos theta_0 frac513, naanzaaligncos theta_n cos 2 theta_n - 1 2 cos2 theta_n - 1 - 1.endalignKwa kuwa mfuatano a_n na cos theta_n una neno la awali sawa, na urejeshaji sawa, unalingana.Tuna hiyosin2 theta_0 1 - cos2 theta_0 frac144169.Kwa kuwa theta_0 ni kali, sin theta_0 frac1213 .Sasa,anzaaligna_0 a_1 dotsm a_n - 1 cos theta_0 cos theta_1 dotsm cos theta_n - 1 cos theta_0 cos 2 theta_0 dotsm cos 2n - 1 theta_0.endalignKuzidisha pande zote mbili kwa sin theta_0 frac1213, tunapataanzaalignfrac1213 a_0 a_1 dotsm a_n - 1 sin theta_0 cos theta_0 cos 2 theta_0 cos 4 theta_0 dotsm cos 2n - 1 theta_0 frac12 sin 2 theta_0 cos 2 theta_0 cos 4 theta_0 dotsm cos 2n - 1 theta_0 frac14 sin 4 theta_0 dotsm cos 2n - 1 theta_0 dotsb frac12n sin 2n theta_0.mwishoalignKwa hiyo,a_0 a_2 dotsm a_n - 1 frac12n cdot frac1312 sin 2n theta_0 le frac12n cdot frac1312.Hii inatuambia c le frac1312.Tunaweza kukokotoa kuwa a_1 2a_02 - 1 2 kushoto frac513 kulia2 - 1 -frac119169, hivyofrac513 cdot frac119169 le fracc4.Kisha c ge frac23802197. The boundfrac23802197 le c le frac1312inatuambia kwamba nambari kamili iliyo karibu na 100c ni 108. 