(7) To find the middle term of (1+x+x2)12. This term will involve a12, and will be found to be =73789x1o. More generally, let it be required to find the middle term of (1 + x + x)". The middle term involves a" and the number n, if broken into parts between 1 and 2 inclusive, will present the following series of fractures. n (n − 1)...2.1, n (n − 1)...2, n. (n − 1)...3 n (n - 1)...4,...... n (n − 1)...(r + 1)...&c. the successive denominators are 1.2.... n, 1.2...(n − 2) × 1., 1.2... (n - 4) × 1.2, 1.2...(n-6)x.1.2.3., ... 1.2...(n − 2r) × 1. 2...r,... (8) 1 + x + x2 + x3)6 = 1 + 6x + 21x2 + 56x3 +120x1 +216x5 + 336x + 456x7 + 546x8 + 586x +546x10+456x11 +336x12 + 216x13 +120x11 +56x15 +21x16+6x17 + x18. 10 14 (9) To find the term of the developement of (1 + x + x2 + x13 + x1)1o which involves 12. The number 12 may be broken into parts, between 1 and 4 inclusive, not exceeding 10 in number, in 31 different ways, the first being 4, 4, 4, and the last 2, 2, 1, 1, 1, 1, 1, 1, 1, 1. The sum of all the terms corresponding to these fractures is 18200512+. is The square of the coefficient of the middle term of (1+x+x2) equal to the sum of the squares of the coefficients of the series for (1+x+x2)" ; and the same relation exists between the middle term of (1+x+x2+.....xTM) and the sum of squares of the coefficients of the series for (1+x+x2+.........ï*)" : for if we make (1+x+x2+..xTM )" =1+a1x+A2 x2 + ... Anm xTMM (1+x+x2+x)=1+a1x+a1⁄2 x2+ɑnm xTM, it will readily appear that the term in their product which involves xam, which is the middle term of (1+x+x2+.......x")", will be = (1 × anm +α1 × Anm−1 +ɑ2 × αam −2+...+Amm−1 × α1 +ɑnm × 1)xTM = (12+a12+a22+. ..... Anm- 12+ anm2) xn, since 1am, A1 = Anm−1, αq=anm-2, and so on, for the coefficients of all the terms which are equidistant from the beginning and the end. In this and similar cases, the term in question may be determined, as well as the developement of the series very easily effected, by considering (1+x+x+x3+x1)1o = (==) 10 and by multiplying together the series for (1-5)10 and (1 — ‚x)—1o ̧ - · (10.) (2 - 5x — 7x2 + x3 + 3x1)5 = 32 - 400x +1440x2+680x3- 11390x1+ 1955x5 + 47025x6 + 5435x7 -1118458-71145x+10807310 +11949511 36185x12 86055x13 816514 +3144115 + 9465x16 - 5715 a17 2565x18+ 405x19 + 243x20. chances. 346. The solution of the following problem is con- Problem in nected with the formulæ and methods we have just been considering : "What is the chance of throwing any assigned number m+n in n throws with one die, or in one throw with n dice?" The faces of the die being marked with the numbers 1, 2, 3, 4, 5, 6, the minimum throw is clearly n, when n aces are thrown. It would be the same problem therefore if the faces of the die were marked with 0, 1, 2, 3, 4, 5, and the number required to be thrown was not m + n but m. The whole number of combinations both favourable and unfavourable is 6" (Art. 245.); but those combinations are alone favourable, the sum of the numbers in which is equal to m: if we represented therefore the several faces of the die by ao, x1, x2, x3, x1, a3, the number of those favourable cases would be that coefficient of the term in the developement of (1 + x + x2 + x3 + x1 + 5)", which involved am: for that coefficient would denote the number of all those combinations of the faces or their representatives which made up the required sum. The chance required will be found by dividing this coefficient by 6". (Art. 266.) Q a It is obvious that the same problem would be solved and collecting those terms in the product of the series for which involved am: the coefficient of this term is as follows; n (n + 1) ... (n + m −1) 1.2.... + ... +(−1)' . Examples. m n. (n + 1) ... (n + m (n+m-7) 1.2... (m − 6) n (n + 1)... (n + m −13) n (n − 1) 1.2. &c. n(n+1)...(n+m−6r−1) n.(n−1)..(n−r+1) 1.2 .. the series terminating, when 6r + 1 is equal to, or greater than, m. 347. To find the chance of throwing 17 in three throws with a single die. Since 17 3 = 14, we must break the number 14 into parts between 1 and 5 inclusive, not exceeding 3 in number: there is only one such fraction, which is 348. To find the chance of throwing 18 in one throw with 5 dice. Since 18 5 = 13, the number 13 must be broken into parts, between 1 and 5 inclusive, not exceeding 5 in number: the number of such fractures is 20, beginning with 5, 5, 3, and ending with 3, 3, 3, 2, 2: the sum of the coefficients of the corresponding terms in the developement of (1 + x + x2 + ∞3 + x2 + ∞3)5 is 780, which denotes the number of cases favourable to the hypothesis made: the chance therefore required The same conclusion may be much more readily deduced from the second formula, which gives us for the favourable cases |