TO THE RESOLUTION OF PROBLEMS. 
125 
Hence, by making d — — and putting the 
values of —— + z m [as expressed in the said problem 68) 
z 
~ Q, and then subtracting above, &c. our equa 
tion becomes dQ — l. + s + s z — 2 + s 3 — 3s + 
s 4 — 4s z + 2, &c. continued to m terms; whence the 
value of s may be determined. 
Thus, let n, the number of terms given, be four; 
then m being =2, Q ( = —j- + z 1 ) will be s z — 2; 
and Our equation will, here, be d x s z — 2=1 + s. 
If n be = 6, Q ( = —+ 2 3 ) will be = s 3 — 3s; and 
we shall have d x s 3 — 3s = l + s + s z — 2 = 
s z 4- s — l; and so in other cases, where n is an even 
number. 
If n be an odd number, put 2m — n — l; and let both 
sides of the equation 
a x 1 + z n — — X 1+2 x l + 2 + 2* . . . z”“ 1 
a 
be divided by 1 + z; so shall 
a X l-2 + 2 2 -2 3 . ..-Z r,-2 +Z n_1 — — V, 1+Z + Z+.. + Z n_1 
a 
(because l + z x 1 — z + z z — z 3 + 2 4 ...—z n ~ 2 + z n_1 
_ 51 — 2 + Z 2 Z 3 + Z 4 ...— 2 71 - 2 + 2”- 1 *) _ 
” l* + z —z 1 + z*—~z+ ... + z*“ 2 — z n ~ l + Z n S ~ 
1 — z n ): whence, by transposition, and substituting m, 
a x 1 + z 2 + z 4 . . . + z 2ra = a + -- X 
a a 
z + z 3 + z s . . . z 2 ™ -1 ; put aa - ^ — c and let the 
r aa — b 
whole equation be divided by n x z m ; then will