Find the sum of n terms of the series whose nth term is 21. Shew that the nth term of the 7th order of figurate numbers is equal to the 7th term of the nth order. 22. If the nth term of the 7th order of figurate numbers is equal to the (n+2)th term of the (r-2)th order, shew that r=n+2. 23. Shew that the sum of the first n of all the sets of polygonal numbers from the linear to that of the 7th order inclusive is SUMMATION BY THE METHOD OF DIFFERENCES. 395. Let un denote some rational integral function of n, and let u1, U2, Uз, U,... denote the values of u when for n the values 1, 2, 3, 4,... are written successively. We proceed to investigate a method of finding un when a certain number of the terms u1, U2, Uз, u,... are given. From the series u1, u2, us, us, u,... obtain a second series by subtracting each term from the term which immediately follows it. The series U2-U1, Uz- U2, Us — U3, U5 - U49... thus found is called the series of the first order of differences, and may be conveniently denoted by which may be called the series of the second order of differences, and denoted by Δ1, Δ2 Δ..... From this series we may proceed to form the series of the third, fourth, fifth,... orders of differences, the general terms of these series being A, Au,, Aur,... respectively. From the law of formation of the series it appears that any term in any series is equal to the term immediately preceding it added to the term below it on the left. Thus u2 = u1 + Au1, and Au2 = Au1 + A„Ù ̧• By addition, since u2 + Au2 = uz we have Ug = W + 2 + Δ.1. In an exactly similar manner by using the second, third, and fourth series in place of the first, second, and third, we obtain So far as we have proceeded, the numerical coefficients follow the same law as those of the Binomial theorem. We shall now prove by induction that this will always be the case. pose that For sup then by using the second to the (n + 2)th series in the place of the first to the (n+1)th series, we have Hence if the law of formation holds for un+1 it also holds for un+2, but it is true in the case of u1, therefore it holds for u,, and therefore universally. Hence 396. To find the sum of n terms of the series in terms of the differences of u1. Suppose the series u1, u2, uз,... is the first order of differences the law of formation is the same as in the preceding article; The formulæ of this and the preceding article may be expressed in a slightly different form, as follows: if a is the first term of a given series, d1, da, dg,... the first terms of the successive orders of differences, the nth term of the given series is obtained from the formula a + (n − 1) d1 + (n − 1) (n − 2) d2+ Example. Find the general term and the sum of n terms of the series 12, 40, 90, 168, 280, 432,...... The successive orders of difference are 28, 50, 78, 112, 152,...... 22, 28, 34, 40,...... 6, 6, 6,...... .... The sum of n terms may now be found by writing down the value of Zn3+52n2+6Σn. Or we may use the formula of the present article and 397. It will be seen that this method of summation will only succeed when the series is such that in forming the orders of differences we eventually come to a series in which all the terms are equal. This will always be the case if the nth term of the series is a rational integral function of n. For simplicity we will consider a function of three dimensions; the method of proof, however, is perfectly general. u1 = An3 + Bn3 + Cn + D, ...... denote the nth term of the first, second, third orders of differences; then that is, v = U2+1 — u2 = A (3n3 + 3n + 1) + B (2n + 1) + C' ; n Thus the terms in the third order of differences are equal; and generally, if the nth term of the given series is of p dimensions, the terms in the pth order of differences will be equal. Conversely, if the terms in the pth order of differences are equal, the nth term of the series is a rational integral function of n of p dimensions. Example. Find the nth term of the series - 1, −3, 3, 23, 63, 129,...... The successive orders of differences are - 2, 6, 20, 40, 66,...... 8, 14, 20, 26,...... 6, 6, 6,...... Thus the terms in the third order of differences are equal; hence we may un=A+Bn+ Cn2 + Dn3, assume where A, B, C, D have to be determined. Putting 1, 2, 3, 4 for n in succession, we have four simultaneous equations, from which we obtain A=3, B=−3, C=− 2, D=1; hence the general term of the series is 3 - 3n — 2n2 + n3. in 398. If a is a rational integral function of p dimensions n, the series is a recurring series, whose scale of relation is (1 − x)p+1. Let S denote the sum of the series; then − a ̧)x2 S(1-x)= a + (a, - a)x+(α, a1) x2 + ... + (α-a) 2" - a_x"+: n = a + b1x + b2x2 + + bx" - ax”+1, say; ... so that b, is of p- 1 dimensions in n. Multiplying this last series by 1 - x, we have here b1 = a α S(1-x) =a ̧+(b,−a )x+(b2-b1)x2+...+(b2-bn-1)x” — (α2+bn)x" +1+α„x”+2 here c1 = b2-bn-1, so that c, is of p - 2 dimensions in n, |