ON POLYGONAL NUMBERS, jg^ 



number is n, =ir 1 -f- w — 2 X « — 1 (Em€r9on*sproporticjn, 

 prop. 6) =m — 2 X + 3 — m; moreover the sum of n 

 terms of the series zzk (by Def* 1) ~z h m — 2 x {• 



3-^m X — (proportion prop. 7) ; therefore A s a + y» — 2 x 



2-^, QED. 



Cor. 1st. Every potygdn of the first order ril ; and every Cor. I, 

 one of the second order r:w; this is proved by substituting 



1 and 2 for a in the expression a -f- m — 2 X ^ =: k. 



Cor. <2d. All polygons of higher orders may be found by Cor. 2. 

 the table in prop. 1st, thus: to the given index ct, add the 



product of the corresponding number — H-, multiplied by 

 m — 2, and the sum is the polygon. 

 Example — Thus the sixth decagon z= 6 + 15 X 8 =: 126. 



Prop. 3d. Let g, h, /, &c. be polygonal numbers of a Vtop. 3. 

 common denomination »?, the indices of which are b, c, d, 

 &c. respectively ; also put b -\- c-\- d, &c. r: a, and let k be 

 the polygon of which the denomination =r?w and index zza; 

 then g -f A + /, &c. + m— 2 X bc + bd+ cd, &c. r: k. 

 For by addition and prop. 2, g -\- h ■{• l, &c. zz b ■}- c -\- d^ 



&c. + ,1^ X *!±|+l!-(;;^2) X *±f:±i; but 



2 2 



fc* + c' 4- d* a' 



' — = — [b c -\- b d ■\- c d) by involution ; 



hence by substitution g ■{- k -^ I :z: a ■]- m — 2 x — 



nc — 2 X6c-l-6d+c£?; therefore g" + A + / 4- ?« — 2 X 



6c + *rf + c</ = a+ w— 2 X - t -=■ k (by prop. 2) 



QED. 



Scholium. Put »i r: 4, and we have (by def. 2 and prop. Scholiufti. 

 2) §•:=&% /* =: cS /= d^ ; therefore ^ + A + / + ?/J — 2 x 

 M 2 if 



