that is n a b c ..., rst... where each of r, s, t... is zero or a positive integer, and The above result can however be at once obtained by the method of Art. 249, as follows. We know [Art. 67] that the continued product (a+b+c+...) (a+b+c+ ...) (a + b + c + ...)... is the sum of all the different partial products which can be obtained by multiplying any term from the first multinomial factor, any term from the second, any term from the third, &c. n-p 8 The term ab c... will therefore be obtained by taking king a from any r of the n factors, which can be done in „Č different ways; then taking b from any s of the remaining nr factors, which can be done in C, different ways; then taking c from any t of the remaining n-r-s factors, which can be done in C different ways; and so on. Hence the total number of ways in which the term a" b c... will be obtained, which is the coefficient of the term in the required expansion, must be n-r-8 Hence the general term in the expansion of (a+b+c+ ...)" is Ex. 1. Find the coefficient of abc in the expansion of (a+b+c)3. Ex. 2. Find the coefficients of a2b2, bcd2 and abcd in the expansion of (a+b+c+d)a. Thus the required coefficients are 6, 12 and 24 respectively. 259. By the previous Article, the general term of the expansion of (a + bx + cx2 + dx3 + ......)" is Hence to find the coefficient of any particular power of x, say of a, in the expansion, we must find all the different sets of positive integral values of r, s, t,... which satisfy the equations s+ 2t+3u+...... = a, r + s + t + u +...... = n. The required coefficient will then be the sum of the coefficients corresponding to each set of values. Ex. 1. Find the coefficient of x5 in the expansion of (1+2x+3x2)a. 14 The general term is 23+2, and the terms required are those for which 8+2t=5 and r+s+t=4. Since each of the quantities r, s and t must be zero or a positive integer, the only possible sets of values are t=2, s = t=1, s = 3, r = 0, the corresponding coefficients being 1, r = 1 and 14 2.32 112 and 14 031 23.3, that is 216 and 96 respectively. Hence the required coefficient is 312. In simple cases the result can be readily obtained by actual expansion. We have (1 + 2x + 3x2)4=1+ 4 (2x + 3x2) +6 (2x+3x2)2 +4(2x+3x2)3 + (2x+3x2)4. Only the last two terms will contain x5 and the coefficients of 5 in these terms will be found to be 216 and 96 respectively, so that the required coefficient is 312. Ex. 2. Find the coefficient of x4 in the expansion of (1+x+x2)3. Ans. 6. Ex. 3. Find the coefficient of x5 in the expansion of (1+x+x2)4. Ans. 16. Ex. 4. Find the coefficient of x8 in the expansion of (2+x - x2)5. Ex. 5. Find the coefficient of x10 in the expansion of (7 + x + x2 + x3 +xa+x5)3. Ans. 0. Ans. 39. Ex. 6. Find the coefficient of the middle term of the expansion of (1 + x + x2 + x3 + x4)5. Ans. 381. 3c, +7c2+11c,+..... + (4n − 1) c2 = 1 + (2n − 1) 2′′. 2 and and = 3C2 +.. {1 + (n + 1) x}(1 + x)"1 = C ̧ + 2C, x+...... + (n + 1)С„". Hence prove that, C ̧2 + 2C ̧2 + 30.3 +. 0 + (n + 1) C23 = (n+2)/2n-1 15. Shew, by expanding {(1+x)" - 1}", where m and n are positive integers, that 17. Shew that, if there be a middle term in a binomial expansion, its coefficient will be even. 18. Shew that the coefficient of x" in the nth power of x2 + (a + b) x + ab is a"+C,3a"-1b+ C ̧3a"-b2 + ... + b". n 1 n α 19. If n be a positive integer and P denote the product of all the coefficients in the expansion of (1+x)", shew that 21. Shew that, if n be a positive integer, (a+b+c+d+e)5 = Σa3 + 5Σa1b + 10Σa3b2 + 20Σa3bc +30Σa2b3c + 60Za2bcd + 120abcde. |