196 The Eev. F. H. Hummel on 



[Incidentally we may note that this rule for the 5th root 

 affords an easy proof of the arithmetical proposition that the 

 final digit of any number is the same as that of its 5th power, 

 and consequently of any (4^> + l)th power, p being any posi- 

 tive integer.] 



I have given this example worked out, with the marginal 

 calculations, to show what such a sum in this method will look 

 like, and to demonstrate the facility of the working. I shall 

 content myself with subjoining rules for the (w + l)th subtra- 

 hend for each of the next half-dozen series. 



6th root. u n+ 1 — u n = S0n 2 (n 2 + 1) + 2. Rule : — multiply n 2 

 by the next higher number, and again by 30, and add 2 + the 

 previous subtrahend. 



7th root. u n+i = 7n(n + l)\n(n + l) + l^ 2 . Take the pro- 

 duct of n and (n + 1), multiply it by the square of the next 

 higher number, and again by 7. But for the higher values of 

 11 the trouble of squaring will lead us to prefer the alternative 

 formula, u n+l — u n = In J 6n 2 (n 2 + 1) + 4w 2 + 2 \ ; multiply n 2 by 

 the next higher number, and again by 6 ; add to the product 

 4 times n 2 and 2 ; multiply the sum by n } and again by 7, and 

 add the previous subtrahend. 



8th root. u n+l -u' n =28n 2 (n 2 + 2)\n 2 + (n 2 + l)\ + 2. Add 

 to n 2 the next higher number, multiply the sum by (n 2 + 2), 

 and again by ?i 2 , and again by 28, and add 2 + the previous 

 subtrahend. 



9th root. u n+l = 3n(n + l)\3(ri 2 + ii + iy + n 2 (n + l) 2 \; a 

 formula which will be more convenient for higher terms in the 

 shape 



3n(?i + l)[n(n + l){3[{n(n + l) + l}{w(w + l) + 2} + l] 



+n ( n+ l)j- + 3]. 



Find the product of n(n + l) ; take the product of the next 

 two consecutive numbers above it, add 1, multiply by 3, add 

 n(ii-\-l); multiply the sum by w(?z + l), add 3, and multiply 

 by 3n(n + l). 



10th root. 



u n+ x -u n = 30n> 2 + 1) { 3n 2 (n 2 + 3) + n 2 + (n 2 + 3) } + 2. 



Take the product of n 2 and (n 2 + 3), multiply by 3, and add to 

 the product n 2 and (n 2 + 3) ; multiply the sum by n 2 , again by 

 (ri 2 + 1), and again by 30, and add 2 + the previous subtra- 

 hend. 



11th root. 

 u n+] = lln(n + l){ii(n + l) + l}[n(n + l){n(?i + l) + l} 



{n(n+l) + 3} + 12. 

 Take the product of n(n+l), multiply it by the next higher 



