Example 1. Find the sum of n terms of the series Example 2. Find the sum to n terms of the series ......9 Here the rule is not directly applicable, because although 1, 2, 3, the first factors of the several denominators, are in arithmetical progression, In this example we may the factors of any one denominator are not. proceed as follows: = + 3 4 (n+2)(n+3)(n+1)(n+2) (n+3) n (n+1)(n+2)(n+3)* Each of these expressions may now be taken as the nth term of a series to which the rule is applicable. 387. In cases where the methods of Arts. 383, 386 are directly applicable, instead of quoting the rules we may always effect the summation in the following way, which is sometimes called 'the Method of Subtraction.' Example. Find the sum of n terms of the series 2.5+5.8+8.11+11.14+ The arithmetical progression in this case is 2, 5, 8, 11, 14,...... ...... In each term of the given series introduce as a new factor the next term of the arithmetical progression; denote this series by S', and the given series by S; then +(3n-1) (3n+2) (3n+5) ; .. S'-2.5.8=5.8.11+8.11.14+11. 14. 17+... to (n-1) terms. By subtraction, -2.5.8-9[5.8+8.11+11.14+... to (n - 1) terms] - (3n - 1) (3n+2) (3n+5), -2.5.8=9[S-2.5]-(3n-1) (3n+2) (3n+5), 9.S= (3n-1) (3n+2) (3n+5) −2.5.8+2,5.9, S=n (3n2+6n+1). 388. When the nth term of a series is a rational integral function of n it can be expressed in a form which will enable us readily to apply the method given in Art. 383. For suppose (n) is a rational integral function of n of p dimensions, and assume $ (n) = A + Bn + Cn (n + 1) + Dn (n + 1) (n + 2) + ... ..., where A, B, C, D,...... are undetermined constants p+1 in number. This identity being true for all values of n, we may equate the coefficients of like powers of n; we thus obtain p + 1 simple equations to determine the p + 1 constants. Example. Find the sum of n terms of the series whose general term is n1+6n3 +5n2= A + Bn + Cn (n + 1) + Dn (n + 1) (n + 2) + En (n + 1) (n+2) (n+3); = − 2, n = - 3 it is at once obvious that A=0, B=0, E1; and by putting n=successively, we obtain C=-6, D=0. Thus na + 6n3 +5n2=n (n+1) (n+2) (n+ 3) − 6n (n + 1). Hence 1 Sn = n (n+1) (n + 2) (n + 3) (n + 4) − 2n (n + 1) (n + 2) 389. If in the expression n + n (n-1)b, which is the sum of n terms of an arithmetical progression whose first term is 1 and common difference b, we give to b the values 0, 1, 2, 3, we get n, 1n (n + 1), n3, 1n (3n -- 1),............. ....... which are the nth terms of the Polygonal Numbers of the second, third, fourth, fifth,......orders; the first order being that in which each term is unity. The polygonal numbers of the second, third, fourth, fifth,......orders are sometimes called linear, triangular, square, pentagonal,...... 390. To find the sum of the first n terms of the rth order of polygonal numbers. The nth term of the 7th order is n + 1n (n − 1) (r − 2); = - 1 n (n + 1) + † (r − 2) (n − 1) n (n + 1) [Art. 383] be taken as the nth term of a new series, we obtain 1, 2, 3, 4, 5, ....... If again we take n (n + 1) which is the sum of n terms of the " 2 last series, as the nth term of a new series, we obtain 1, 3, 6, 10, 15, By proceeding in this way, we obtain a succession of series such that in any one, the nth term is the sum of n terms of the preceding series. The successive series thus formed are known as Figurate Numbers of the first, second, third, ... orders. 392. To find the nth term and the sum of n terms of the rth order of figurate numbers. The nth term of the first order is 1; the nth term of the second order is n; the nth term of the third order is Σn, that is In (n + 1); the nth term of the fourth order is Σ n (n + 1)(n+2). 1.2.3 that is that is n (n + 1) ; the nth term of the fifth order is Σ n (n + 1) (n + 2) (n+3) 14 ; and so on. Thus it is easy to see that the nth term of the 7th order is Again, the sum of n terms of the 7th order is n (n + 1) (n + 2) ... (n + r− 1) NOTE. In applying the rule of Art. 383 to find the sum of n terms of any order of figurate numbers, it will be found that the constant is always zero. 393. The properties of figurate numbers are historically interesting on account of the use made of them by Pascal in his Traité du triangle arithmétique, published in 1665. The following table exhibits the Arithmetical Triangle in its simplest form Pascal constructed the numbers in the triangle by the following rule : Each number is the sum of that immediately above it and that immediately to the left of it; thus 15=5+10, 28 = 7+ 21, 126 = 56 + 70. From the mode of construction, it follows that the numbers in the successive horizontal rows, or vertical columns, are the figurate numbers of the first, second, third,... orders. A line drawn so as to cut off an equal number of units from the top row and the left-hand column is called a base, and the bases are numbered beginning from the top left-hand corner. Thus the 6th base is a line drawn through the numbers 1, 5, 10, 10, 5, 1; and it will be observed that there are six of these numbers, and that they are the coefficients of the terms in the expansion of (1 + x). The properties of these numbers were discussed by Pascal with great skill: in particular he used his Arithmetical Triangle to develop the theory of Combinations, and to establish some interesting propositions in Probability. The subject is fully treated in Todhunter's History of Probability, Chapter II. 394, Where no ambiguity exists as to the number of terms in a series, we have used the symbol Σ to indicate summation; but in some cases the following modified notation, which indicates the limits between which the summation is to be effected, will be found more convenient. Let (x) be any function of x, then Σ (x) denotes the sum of the series of terms obtained from (x) by giving to x all positive integral values from 7 to m inclusive. For instance, suppose it is required to find the sum of all the terms of the series obtained from the expression (p-1) (p-2)... (pr) by giving to p all integral values from r+ 1 top inclusive. H. H. A. 21 |