EXAMPLES FOR EXERCISE. 1. How many more balls are there in a square pile of 15 courses than in a triangular one? Ans. 560. 2. A square pile which has as many courses as a triangular pile, contains half as many more balls. How many balls were there in both? Ans. 4 courses in each, and 50 balls in all. 3. The upper and lower courses of an incomplete square pile have 15 and 25 balls in each side. - Find the number in the original pile, and the number left. Other examples may be easily formed to suit the ability of the pupil. THE BINOMIAL THEOREM. This theorem affirms that every expression of the form (a + x)" can be developed in a series of positive integer powers of either a or x, and assigns the coefficients of those powers of a or x in the development. It does not, however, affirm that this is the only form of development possible. The expression is according as we consider a or x the leading term of the given expression. 1. To find the first term of the development. .... Assume (a + x)” = An + B2x + С2x2 + D„x3 + (a) Then, using similar notation where 2n is written for n, we have (a + x)2n = A2n + B2nx + С2nx2 + D2nx3 + (b) But (a + x)2= {(a + x)"}2, in which, substituting from (a) and (b), we get whence, equating the coefficients of ao, we get A„2 = A2: which is fulfilled by A, a", and the first term is found to be a" universally. 2. To find the second term of the development. It will simplify the process to write the expression (a + x)” = a” ~ { 1 + 2 } * = a"(1 + v)"; and it will evidently be sufficient to expand the compound factor, and multiply every term of the expansion by a" to obtain the complete develop ment. First, let n be a positive integer: then we may assume as before, (1 + v)” = 1 + A„v + B„v2 + C„v3 + .... ...... ...... (c). For since a" {1 + A„v + } gives a + a"A„v + the first term of the bracketted series being 1, fulfils the condition respecting the development deduced in the foriner part of the investigation. Divide, synthetically, equation (c) continually by 1+v: then (1 + v)”—” = 1 + (A„ − m) v + Bn-m v2 + Cn–m v3 + and if we take m=n-1, we have .... (n-1), or A, = n. Equating the coefficients of the homologous powers of v, we have for those of the first powers, or v1, When, therefore, n is a positive integer, A,= n, and the coefficient of the second term is found. Secondly, let n be a negative integer; or the expression to be developed be (1 + v)—". This is the same thing as 1 = 1 (1 + v)" 1 + nv + B2 v2 + = 1−nv + B_n v2 + С_n v3 + .... by actual division. Whence in this case we have (1 + v)−* = 1 − nv + and the coefficient of the second term is found. Thirdly, let n be a fraction denoted by ± (1+0) i = 1 + A2 v + B2 v2 + C2 v3 +.... Raise both sides to the qth power: then we have 1±pv+B, v2 + ...=1+qv (A,+B„ v+... . .) + B2 v2 (An+ B2 v+..)2 + .... and equating the homologous coefficients of v, we have for that of the first power, ±p=qA,, or A2 = ± 2. Whence in this case also the coefficient of the second term is found: and in all cases it is equal to the index of the power. 3. To find the third and subsequent terms of the development. Put vy+2: then we have, denoting the coefficient of (y + z)" by A, throughout, (1+y+2)=1+n(y + z)+A2 (y+2)2+As (y+2)3 + ... A,+1 (y + z) '+1+... Now by indeterminate coefficients, the coefficients of each separate power of z in these two developments are equal. Equate them for z1: then n + 2A, y + 3A ̧ y2 + ... (r + 1) A‚+1y. + = (1 + y)" or multiplying out, and changing sides, n(1 + y)" = (1 + y) {n + 2A, y + 3A, y2 + ... (r + 1) A‚ + 1 3′′ + ... ... ... . } ; or again reducing, the two sides of the equation become respectively n{1+ny +A, y2 + .... A, y2 + ...} and (n+(2A2+n) y+(3A ̧+2A,) y2+(4A, +3A,) y3 + .... {(r+1) A,+1+rA,}y'+... = : (n − r) A,. Again, equating the homologous coefficients in these two expansions, we get, generally, nA‚=rA, + (r + 1) A, + 1, or (r + 1) A, + 1 Substituting in this the successive values of r, we have rA, = {n — (r — 1)} A‚-1 where, by the preceding investigation of the second term, we have A。 = 1. Multiply these columns vertically, and there results of v; and giving 1.2.3 ....r which is the general form for the coefficient of the th power tor the successive values, 1, 2, 3, we have the corresponding coefficients, .... the forms assigned in the enunciation of the theorem. Inserting these, restoring the value of v, viz., and multiplying α all the terms by a", we have the formula as there given. Thus we obtain Similarly, since a − x = a + (− x), we have in the expansion all the odd n powers of a negative; that is, {a + (— x)}" = (a — x)" = a* — — a′′ a"-1 x + 1 4. When ʼn is a positive integer, this series terminates with the (n + 1)th term, and the coefficients of the terms reckoned from either extremity of the series are equal. In all other cases the series will never terminate. For the (n+2)th term contains the factor 0, and this factor entering into all the succeeding terms, they also become 0. Whence the series terminates with the (n+1)th term. Again, the 7th term from the end is the (n -r+2)th from the beginning; and therefore its coefficient is n(n − 1) (n ... 2) r ·r + 1) and the coefficient of the 7th term from the beginning is .... It, therefore, only remains to show that PQ. Now we have very obviously (n − r + 2) in which the numerator and denominator are manifestly equal. as above affirmed. r + 2) (n - 1) n Hence P = Q, 5. It still remains to point out the arithmetical forms which are most convenient in the practical application of this theorem. Suppose, as an instance, we had to develop (a + x)±10, the calculations will be as follow: Write the terms without coefficients a10 a9x between to receive the coefficients and their signs. a8x2 aa.... with spaces Then when, and so on to the required extent: the literal parts of this throughout being merely explanatory, and need not be put down in actual working. The process is simply multiplying by the decreasing series, and dividing by the increasing one alternately. Each successive quotient is the successive coefficient of the series, which inserted in its place, gives the expansion sought. This vertical alineation is not, however, convenient when n is a fraction, a horizontal one being much preferable. Thus, to expand (1 + v), we may continually work as follows: where the asterisks are placed at the several successive coefficients. The apparent continuity of equality may be, were it necessary, cut off, by drawing vertical lines after each sign of equality that is to be destroyed by the next operation, as in the example above. This cutting off is better than crossing out: but neither of them is absolutely necessary. EXAMPLES IN POSITIVE INTEGER POWERS. Ex. 1. Raise a - to the 10th power. Ex. 2. Find the sixth power of a—x. Ex. 3. Find the fourth power of a ― x. Ex. 4. Involve a— x to the ninth, and a + b − c to the fourth power. Ex. 5. Find the cube of a- √b. Ex. 6. Find the fifth term of (3y — 2x)9. Ex. 7. Raise - ab to the fourth power. -b. EXAMPLES OF NEGATIVE AND FRACTIONAL POWERS. Ex. 1. Extract the square root of a2 + b2 in a series; or evolve (a2 + b2)3⁄4. b2 b4 b6 + 568 Ans. a + 2α 8a3 16a5 128a7 Ex 12. Assign the first eleven coefficients of (1 + 2)1 +*. }, where √b—c=d. √b-c. 935d8 Ex. 14. It is required to find the square and the cube of the expression √x-y√ √ −x + y √ − 1. -X Scholium. The binomial development may be employed also in the extraction of roots of numbers, and sometimes with considerable advantage. It is especially the case when the number whose root is to be extracted does not differ greatly from the same power of some whole number, as in such cases the convergency is so rapid as to give eight or ten figures of the root true, by means of three or four terms of the development. * Examples 3, 5, 9, may be verified by Synthetic Division; and in all cases where it can be applied, and the result merely is required, this method is rather simpler than the binomial theorem. |