Putting My2 for all the terms after By, we have y + (1 + y)" = 1 + By + My, n log (1+ y) = log (1+ By + My), Art. XXVII., and, if we assume log (1+ y) = a + by + cy2 + &c, a, b, c, &c., being independent of y; then ..nanby+ncy2+ &c. = a+bBy+bMy2+ &c., .. by Art. XXVI., na = a, ..a = 0, nb = b B, ... B= n. Hence, substituting n for B in (4), we have (1 + y)" = 1 + ny + n (n 1) 1.2 n (n − 1) (n 2) y3 + &c., 1) -2 a"-2x2 + n (n − 1) (n − 2) = a" + na"-1 x + which is the Binomial Theorem of Sir Isaac Newton. Cor. 1. If a be negative, its even powers will be positive, and its odd powers negative; hence Cor. 2. If a = 1, (1 + x)" = 1 + n x + ·.· (a + b + c)" = and (a+b+c + d)" = {(a + b) + c}", {(a + b) + (c + d)}", the Binomial Theorem may be applied to the expansion of powers of a polynomial. The above proof is general, and consequently applies to the cases of n being an integer, or a fraction, positive or negative. On inspecting this theorem, it appears that The powers of a decrease and those of x increase by unity in each succeeding term. The first term is the nth power of the first term of the given binomial. The second term is found by multiplying the index of the first by the first with its index diminished by unity, and also by the second term of the binomial. The coefficient of any succeeding term is found by multiplying the coefficient and index of a in the preceding_term together, and dividing by the number of terms already set down. In applying this theorem, it will appear that When n is an integer, the number of terms in the expansion is n + 1; and the coefficients of any two terms equidistant from the extremes are equal to each other. When n is a fraction the series does not terminate. = a + 6a5x + 15 a2x2 + 20 a3 x2 + 15 a2x2 + 6 ax3 + xo. Practically, the coefficients after the second term are found and after the middle term they may be written in an inverse order. = 2. (a — x)5 — a3 — 5 a1x + 10a3x2 - 10 a2x2 + 5a x1 —- x3 3. (1-x)=1-7x+ 21 x2—35 x3 +35 x1—21 x2+7 xˆ —x3. 4. (2x + 1)2 = 32x2 + 80x1 + 80x3 + 40x2 + 10x + 1. Hence, by the Binomial Theorem (Cor. 2), a2 = { 1 + n (a − 1) + n(n-1) (n-2) 2.3 n (n − 1) 2 (a1)2+ (a − 1)* + &c. }* = = where B, C, &c. contain powers of (a — 1) only. x (x n) 1 + x(A + B n + &c.) + (A + Bn + &c.)2+ &c. 2 = 0, then Now since n may have any value whatever, leɩ n = If e be that value of a which makes A = 1, the base of the system of logarithms used by Napier, the inventor of logarithms. VARIATIONS, PERMUTATIONS, AND COMBINATIONS. XXXI. The different arrangements which can be made of any number of quantities, taking a certain number at a time, are called their Variations. Thus, if a, b, c be taken two together, their variations will be ab, ba, ac, ca, bc, cb. If all the quantities are taken together, their variations are called Permutations. Thus the permutations of a, b, c are abc, acb, bac, bca, cab, cba. XXXII. The number of variations of n different things, taken r together, is n (n − 1) (n − 2) . . . . {n — (r — 1)}. Let a, b, c, d, &c., be the n things; then the number of variations which can be made, taking them singly, is n. Let n 1 of these things, namely b, c, d, &c., be taken singly, then the number of their variations is n - 1: and if a be placed before each, we shall have n 1 variations of n things, taken two together, in which a stands first; similarly we shall have n - 1 such variations, in which b stands first; and similarly for all the n things, hence there will be on the whole n (n-1) variations of n things taken two together. Again, taking n-1 of these things, namely b, c, d, &c., their variations, taken two together, will be n (n 1) (n and proceeding as before, there will be on the whole (n - 1) (n - 2) variations of n things taken three together. 2); Similarly their variations, taken four together, will be n(n - 1) (n-2) (n-3). Hence, if V, V, V, &c., V, denote the variations of n things, taken 1, 2, 3, &c., r together, we have ‚ = n (n − 1) (n − 2) (n − 3).... {n (r− 1)}. Cor. the permutations (p) of n things are their variations, taken all together; by writing n for r we shall have p = n(n − 1) (n − 2) . . . . {n − (n − 2)} {n − (n − 1)} -- .... XXXIII. When a and b are different, their permutations are ab, ba, but when a = b, they become aa. Let a recur p times. b, q times, c, r times, &c., and P be the number of permutations required. Then if all the a's be changed into different letters, they will form 1.2.3.. ....P |