⅓ b × px + ⅓ b × qx + ⅓ b × rx = 0
⅓ c × p²x² + ⅓ c × q²x² + ⅓ c × r²x² = 0
⅓ d × p³x³ + ⅓ d × q³x³ + ⅓ d × r³x³ = dx³
⅓ e × p⁴x⁴ + ⅓ e × q⁴x⁴ + ⅓ e × r⁴x⁴ = 0 &c.
And consequently,
p + q + r = 0
p² + q² + r² = 0
p³ + q³ + r³ = 3
p⁴ + q⁴ + r⁴ = 0, &c.
Make, now, p³ = 1, q³ = 1, and r³ = 1; that is, let p, q, and r, be the three roots of the cubic equation z³ = 1, or z³ - 1 = 0: then, seeing both the second and third terms of this equation are wanting, not only the sum of all the roots (p + q + r) but the sum of all their squares (p² + q² + r²) will vanish, or be equal to nothing (by common algebra), as they ought, to fulfil the conditions of the two first equations. Moreover, since p³ = 1, q³ = 1, and r³ = 1, it is also evident, that p⁴ + q⁴ + r⁴ (= p + q + r) = 0, p⁵ + q⁵ + r⁵ (= p² +q² + r²) = 0, p⁶ + q⁶ + r⁶ (= p³ + q³ + r³) = 3. Which equations being, in effect, nothing more than the first three repeated, the values of p, q, r, above assigned, equally fulfil the conditions of these also: so that the series arising from the addition of three assumed ones will agree, in every term, with that whose sum is required: but those series’ (whereof the quantity in question is composed) having all of them the same form and the same coefficients with the original series a + bx + cx² + dx³, &c. (= S), their sums will therefore be truly obtained, by substituting px, qx, and rx, successively, for x, in the given value of S. And, by the very same reasoning, and the process above laid down, it is evident, that, if every nth term (instead of every third term) of the given series be taken, the values of p, q, r, s, &c. will then be the roots of the equation zⁿ - 1 = 0[155]; and that, the sum of all the terms so taken, will be truly obtained by substituting px, qx, rx, sx, &c. successively for x, in the given value of S, and then dividing the sum of all the quantities thence arising by the given number n.
The same method of solution holds equally, when, in taking every nth term of the series, the operation begins at some term after the first. For all the terms preceding that may be transposed, and the whole equation divided by the power of x in the first of the remaining terms; and then the sum of every nth term (beginning at the first) will be found by the preceding directions; which sum, multiplied by the power of x that before divided, will evidently give the true value required to be determined. Thus, for example, let it be required to find the sum of every third term of the given series a + bx + cx² + dx³ + ex⁴, &c. (= S), beginning with cx². Then, by transposing the two first terms, and dividing the whole by x², we shall have c + dx + ex² + fx³, &c. = S - a - bx ⁄ xx (= S´). From whence having found the sum of every third term of the series c + dx + ex² + fx³, &c. beginning at the first (c), that sum, multiplied by x², will manifestly give the true value sought in the present case.
And here it may be worth while to observe, that all the terms preceding that at which the operation (in any case) begins, may (provided they exceed not in number the given interval n) be intirely disregarded, as having no effect at all in the result. For if in that part (- a - bx ⁄ xx) of the value of S´, above exhibited, in which the first terms, a and bx, enter, there be substituted px, qx, rx, successively, for x (according to the prescript) the sum of the quantities thence arising will be
- a ⁄ p²x² - a ⁄ q²x² - a ⁄ r²x²
- b ⁄ px - b ⁄ qx - b ⁄ rx
which, because p³ = 1, q³ = 1, &c. (or p² = 1 ⁄ p, q² = 1 ⁄ q, &c.) may be expressed thus;
- a ⁄ xx × (p + q + r)
- b ⁄ x × (p² + q² + r²)
But, that p + q + r = 0, and p² + q² + r² = 0, hath been already shewn; whence the truth of the general observation is manifest. Hence it also appears, that the method of solution above delivered, is not only general, but includes this singular beauty and advantage, that in all series’ whatever, whereof the terms are to be taken according to the same assigned order, the quantities (p, q, r, &c.), whereby the resolution is performed, will remain invariably the same. The greater part of these quantities are indeed imaginary ones; and so likewise will the quantities be that result from them, when substitution is made in the given expression for the value of S. But by adding, as is usual in like cases, every two corresponding values, so resulting together, all marks of impossibility will disappear.