(becaufen f is n ;) therefore n +n2 4.16 If n be an indefinite Number, and be 1, then {1* + 2o +33 + &c. +2 is (by what has been said in÷) 22 p+2 x n 4.) Again, if p be=2, then 12+ +2+3 + &c. + n + 'will be (by our Lem, 2.) •, &c. Corollary-I. + Wherefore if n be an indefinite Number, and p = any Affirmative whole Number 1, 1+ 2+ 3+ &c. + m2 And confequently 1 + 22 + 32 + &c. + 2 Scholium II. 1. Again, by Scholium to Prob. 2. Chap. 3. of this part 3. it is further demonftrated, that p being Number 1+2" + 3" + &c. +n" is= I t any affirmative whole + I + fore? fore I our Lemma 2.) = &c. 1 +3 + 3 2. Likewife by the faid Scholium to Prob. 2. Chap. 3. of this I Part 3. it is demonstrated that 1' + 2' + 3' + &c. continu'd xx n'; then the Sum of this Series continu'd to n Terms +&c. continu'd to n Terms, will be (by our Lemma 2.) of the laft Series continu'd only to n Terms is From what has been faid it is manifeft, that q being likewife any affirmative whole Number 22 continu'd to{"} Terms is = "+" + 1 P 5.2. {:} Wherefore it may be demonftrated by the like Method with that us'd in Prob. 2. and its Sch. in Chap. 3. of this Part 3. Now, tho' p and q be limited fo as to be equal to whole Numbers, yet there is no Number either whole or fracted If m be any affirmative Number whatsoever, and n = an Indefinite Number 1+2+3+4+ &c. continu'd VA= x; then AB the Property of a Circle Given: 2rx, and Required to find AEt. Note, AD LVB. Preparation. Produce the Tangent ED to H, and imagine the Line HGF to be 1, and indefinitely - near DA, and LKI and indefinitely near DA, and (if you will) let AI be AF 4. Solution. By the Property of a Circle A B x VA□ DA; that is, 2r-xxx=2rx-xx=DAg: and FBxVF=OFG; that is, 2rxaxx + a = 2rx + 2ra - x2 - 24x 44 FGq: The The As EHF and EDA are fimilar; wherefore EA EAq: EFq::: ADg: FHq: That is, to, or But FHFG (let a be ever so small, unless it be lefs than nothing, as it is not by Suppofition;) therefore.. FHq: FGq; 1 The foregoing Step xtt, produces 2 rætt +4 г xì å +2rxaa- x xtt 2 txxa -x xtt - 2axt t 2 txxa 2 2 - a2x2 2 ttra — 2ttax aaxx 2rxt t + 2 ratt aatt: Therefore 4rxta + 2rxa a ttaa: And by.. dividing each part by a, and tranfp. 4rxt + 2rxa + tt a - xoac 2ttr 2 t t x + 2 txx: But 2 rxaxxa, and confequently 2 rxa+tta muft bexa; Wherefore the foregoing Step may be defign'd thus, 4rt2ttr-2 ttx +2txx. 2dly, IB XVI KIq, that is 2r-x+ax x — a xx + 2 axaa KIq. =27x - 2ra 良 The AEIL and E A D are Similar; wherefore E A.. 2 ratt x22 + 2 axtta att; confequently 2 tx x + 2 r t t − 2 t + x + 2 r xa+tta-x2a = 47xt; that is 2 tx * +2rt2 - 212x4rxt. That is, in Words, any Quantity whatsoever, that is more than nothing, being added to either of the Quantities 4 r x t, or 2 t x x + 2 t tr 2 tt x, makes the Sum greater than the other Quantity; confequently by* our Lemma 1.4rxt=2tx x + 2 t t r 2 ttx; whence 4rx 2 x x = 2 tr 2tx: Finally, by |