Scholium I. If n be an indefinite Number, and p=1; then 12 + 22 + 32 + &c. + n2 is (by what has been faid in -:-) +I + 3P+1 + &c. +; confequently 1P+1+2+1 + 3P. 2 + P +I X 12 ). Again, if pbe=2, then 1P 2P + 3? +2 + 2P + x + 3 + x +&c.+2 P+1 will be (by our Lem. 2.) •; &c. Corollary I. Wherefore if n be an indefinite Number, and any = p = affirmative whole Number 1, 1P + 2P + 3o + &c. +n2 +1⁄21⁄23 •: And confeqently 1+ 22 + 3o will be 1. Again, by Scholium to Prob. 2. Chap. 3. of this Part III. it is further demonstrated, that ʼn being an indefinite Number, and any affirmative whole Number, 1' + 2 + 2. Likewife, by the faid Scholium to Prob. 2. Chap. 3. of this Part III. it is demonftrated that 1+&+3 + &c. continued 2 72 P +2 + I + 3' + &c. continued to n Terms, will be (by our Lem From what hath been faid, it is manifeft that 4 being like wife any affirmative whole Number, 1 + &c. = 1 + 2 +1 P 3 ed to {22-15 n n Ꮻ . &c. +2 +2 -+ may be demonftrated, by the like Method with that us'd in Probl. 2. and its Schol. in Chap. 3. of this Part, that Now tho' p and q be limitted fo as to be equal to whole Numbers, yet there is no Number, either whole or fracted, but Corollary II. Ifm be any affirmative determinate Number whatfoever, and an indefinite Number, I m m m + 2 + 3 + 4 +&c. continued to 72, and: 22 1: Terms will be equal to 92m m+1 'Tis required to find the Sub-Tangent A E=t=? Draw A DUB; draw alfo the Tangent ED, which produce to H, and imagine the Lines HGF and LK Í, interfecting the Tangent produc'd in H and L, the Arc in G and K, and the Diameter in F and I, to be drawn Parallels to, and at an equal indefinitely fmall Diftance from DA, and suppose that Distance to be a=AFAI. Solution. By the Property of a Circle B A x VA 30: X x0 = 2rx xx; alfo BF x UF UI IK4; that is, DAq:27 — FG4, and BI × :2rxa:x:x+a:=2rx-xx+2ra 2 x A = a a= FG4 1=1Kq The As EDA, EHF and ELI are fimilar; wherefore But FHq is FGq, and ILqIKg (let a be ever so fmall, unless it be equal to, or lefs than Nothing, as it is not by Suppofition); that is, 2rx-xx +4rxta + 2 x x ta + tt ±zra+2xa-aa. And, by Abbreviat. Divif. Multiplicat. and Tranfpofit. +4rxt 2xx t + 2 rx a + tt a 2 x tt. xx a = ± 2 rtt + That is to fay, +4rxt → 2 x x t +: 2rx+tt-x x -2xxt+: :xa+2rtt 2 x tt. And-4rxt+2 x x t + : 2 rx + tt - xx:xa2 rtt + 2 x tt; Confequently, (from the laft Step) 4rxt — 2 x xt — : 2 rx + tt- xx:xa+2rtt. 2 xtt. From the laft, and laft but two Steps, it is manifeft that any Quantity, be it ever fo fmall, fo that it is more than Nothing, (viz. 2rx + tt-xx:x a) being taken from or added to 4rxt 2xxt gives the Remainder or Sum accordingly lefs or greater than 2rtt-2xtt; confequently (by our * Lemma 1.) 4rxt 2xxt=2rtt 06 00 =t. II. • Lema 1. to Ex hauftions. r-x PROP. Part III. and LKI, interfecting the Tangent produc'd in H and L, the Curve in G and K, and the Abfciffe produc'd in F and I, to be drawn |s to, and at an equal indefinitely fmall Distance from DA; and fuppofe that Distance to bea=A FAI Solution. By the Property of a Parabola px= A Dq; as alfo px + GF4 pa= KIq The As TAD, TFH and TIL are fimilar; therefore STFqFHq +2pxta+px a a tt That is ttp::tt+zta+aa ̈px: +2x1+xatt; that is to say, 2x trai tạt xatt: And 2xtxa - tt: + 2 x t-xa+tt. Now from the laft, and laft but two Steps, 2 xt is (by our Lemma 1.) tt: Confequently, 2 x = t. "Tis required to find the Sub-Tangent PT=t=? Draw PM an ordinate to A B; draw alfo the Tangent TM, which produce to H, and imagine the Lines HGF and LKI, interfecting the Tangent produc'd in H and L, the Curve in G and K, and the tranfverfe Diameter in F and I, to be drawn at |