We may observe that the expansion of a binomial can always be reduced to the case in which one of the two quantities is We may then expand (1+y)" and multiply each term by x", and thus obtain the expansion of (x+a)". 513. To find the sum of the coefficients of the terms in the expansion of (1+x)”. 514. The sum of the coefficients of the odd terms in the expansion of (1+x)" is equal to the sum of the coefficients of the even terms. Put x=-1 in the expansion of (1+x)"; thus n(n-1) n (n − 1) (n − 2) 0 = 1− n + 1.2 1.2.3 + ...... = sum of the odd coefficients. sum of the even coefficients. Since then the sums are equal, by the preceding article each 515. The result in Art. 513 gives a theorem relating to Combinations. For suppose there are n things; then we can take them singly in n ways, we can take them two at a time n (n − 1) in 1.2 ways, we can take them three at a time in n(n-1) (n-2) 1.2.3 ways, and so on. Hence by Art. 513 the total number of ways of taking.n things is 2′′ – 1. This theorem was obtained by the early writers on Algebra before the Binomial Theorem was known; the proof is a simple example of mathematical induction which is deserving of notice. We have to shew that if unity be added to the total number of ways of taking n things, the result is 2". Suppose we have four letters a, b, c, d; form all the possible selections and prefix unity to them. Thus we have 1, a, b, c, d, ab, ac, ad, bc, bd, cd, abc, abd, acd, bed, Here the total number of symbols is 16, that is, 24. Now take an additional letter e; the corresponding set of symbols will consist of those already given, and those which can be formed from them by affixing e to each of them. The number will therefore be doubled; that is, it will be 25. The mode of reasoning is general, and shews that if the theorem is true for n things, it is true for n + 1 things. EXAMPLES OF THE BINOMIAL THEOREM. 1. Write down the 3rd term of (a + b)1. 4. Write down the 2001st term of (aio + X10) 2002 6 8. Write down all the terms of (5 9. Write down the middle term of (a + x)1o. 10. Write down the two middle terms of (a + x)o. 11. Expand {a + √(a3 − 1)}® + {a −√(a2 - 1)}° in powers of a. 12. Write down the coefficient of y in the expansion of 13. y If A be the sum of the odd terms and B the sum of the even terms in the expansion of (x + a)", prove that +1 A2- B2 = (x2-a3)". 14. Prove that the difference between the coefficients of x+1 and x" in the expansion of (1 + x)"+1 is equal to the difference between the coefficients of x*+1 and x ̃ ̄1 in the expansion of (1 + x)". 1 1 15. Shew that the middle term in the expansion of (1 + x) 2m 16. Find the binomial expansion of which four consecutive terms are 2916, 4860, 4320, 2160. Prove that the coefficient of x in the expansion of 2r+1 18. Write down the coefficient of x+1 in the expansion of 19. Find the 7th term from the beginning, the 7th term from the end, and the middle term of (x ין 2n+1 20. If to, t,, t t...... represent the terms of the expansion of (a + x)", shew that (t − t + t - ... ... )2 + (t, — ty + t ̧ - ... ... ) * = (a® + x®)". XXXVI. 516. BINOMIAL THEOREM. ANY EXPONENT. We have seen that when n is a positive integer We now proceed to shew that this relation holds when n has any value positive or negative, integral or fractional, that is, we shall prove the Binomial Theorem for any exponent. We shall make some observations on the proof after giving it in the usual form. 517. Suppose m and n are positive integers; then we have hence the product of the series which form the right-hand members of (1) and (2) must = (1 + x)"+"; that is, Equation (3) has been proved on the supposition that m and n are positive integers; but the product of the two series which occur on the right-hand side of (3) must be of the same form whatever m and n may be; we therefore infer that (3) must be true whatever m and n may be. We shall now use a notation that will enable us to express (3) briefly. Let f(m) denote the series whatever m may be; then f(n) will denote what the series becomes when n is put for m; and ƒ(m +n) will denote what the series becomes when m+n is put for m. And when m is any positive integer f(m) = (1+x)"; also f(0) written Similarly, f (m+n)=f(m) xf(n) = 1. Thus (3) may be f(m+n+p) = f(m + n) ׃(p) =ƒ (m) ׃ (n) ׃(p). Proceeding in this way we may shew that f(m + n + p + q + ......) =ƒ (m) × ƒ (n) × ƒ (p) ׃ (q) × ...... (4). (5). Now let m=n=p=q= ==, where s and r are positive ...... integers, and suppose the number of terms to be r; then (5) But since s is a positive integer ƒ (8) = (1 + x), and therefore therefore This 8 proves positive quantity. |