Newton's Binomial Theorem Legally Demonstrated by Algebra. By the Rev. William Sewell, A. M. Communicated by Sir Joseph Banks, Bart. K. B. P. R. S.

XVI. Newton's Binomial Theorem legally demonstrated by Algebra. By the Rev. William Sewell, A. M. Communicated by Sir Joseph Banks, Bart. K. B. P. R. S. Read May 12, 1796. Let \( m \) and \( n \) be any whole positive numbers; and \( 1 + x^n \) a binomial to be expanded into a series, as \( 1 + Ax + Bx^2 + Cx^3 + \ldots \), &c. where \( A, B, C, D, \ldots \) are the coefficients to be determined. Assume \( v = 1 + x^n = 1 + Ax + Bx^2 + Cx^3 + Dx^4 + \ldots \), &c. And \( z = 1 + y^n = 1 + Ay + By^2 + Cy^3 + Dy^4 + \ldots \), &c. Then will \( v = 1 + x \), and \( z = 1 + y \ldots v - z = x - y \). And \( v - z = A \times x - y + B \times x^2 - y^2 + C \times x^3 - y^3 + D \times x^4 - y^4 + \ldots \), &c. Consequently \( \frac{v - z}{v^n - z^n} = A + B \times x + y + C \times x^2 + xy + y^2 + D \times x^3 + x^2y + xy^2 + y^3 + \ldots \), &c. Now \( v - z \times : v^{n-1} + v^{n-2}z + v^{n-3}z^2 + \ldots z^{n-1} \). Also \( v - z = v - z \times : v^{n-1} + v^{n-2}z + v^{n-3}z^2 + \ldots z^{n-1} \). Therefore \( \frac{v - z}{v^n - z^n} \) reduces to, and becomes \( \frac{v^{n-1} + v^{n-2}z + v^{n-3}z^2 + \ldots z^{n-1}}{v^{n-1} + v^{n-2}z + v^{n-3}z^2 + \ldots z^{n-1}} \). \( = A + B \times x + y + C \times x^2 + xy + y^2 + D \times x^3 + x^2y + xy^2 + y^3 + \ldots \), &c. The law is manifest; and it is likewise evident that the numerator and denominator of the fraction, respectively terminate in \( m \) and \( n \) terms. Suppose then \( x = y \); then will \( v = z \); and our equation will become \( \frac{mv^n}{nv^m} = A + 2Bx + 3Cx^2 + 4Dx^3 + \ldots \), &c. But \( v^n = 1 + x \), therefore by multiplying we have \( \frac{mv^n}{n} = A + A + 2Bx + 2B + 3Cx^2 + 3C + 4Dx^3 + \ldots \). Or \( v^n = \frac{1 + x^n}{1 + x^n} = \frac{nA}{m} + \frac{nA + 2nB}{m} x + \frac{2nB + 3nC}{m} x^2 + \frac{3nC + 4nD}{m} x^3 + \ldots \), &c. Compare this with the assumed series, to which it is similar and equal, and it will be \[ \begin{align*} \frac{nA}{m} &= 1 \\ \frac{nA + 2nB}{m} &= A \\ \frac{2nB + 3nC}{m} &= B, \\ &\ldots, &\ldots. \end{align*} \] \[ \therefore A = \frac{m}{n}; \quad B = \frac{m - nA}{1.2.n}; \quad C = \frac{m - 2nB}{1.2.3.n}; \quad \ldots \] Therefore \( \frac{1 + x^n}{1 + x^n} = 1 + \frac{m}{n} x + \frac{m \times m - n}{1.2.n^2} x^2 + \frac{m \times m - n \times m - 2n}{1.2.3.n^3} x^3 + \ldots \), &c. The law is manifest, and agrees with the common form derived from other principles. Sch. In the above investigation, it is obvious that unless \( m \) be a positive whole number, the numerator abovementioned does not terminate: it still remains, therefore, to shew how to derive the series when \( m \) is a negative whole number. In this case, the expression \( (v^n - z^n) \) assumes this form, \( \frac{1}{v^n} - \frac{1}{z^n} \), or its equal \( \frac{z^n - v^n}{v^n z^n} \), which divided by \( v^n - z^n \), as before, gives \[ \frac{1}{v^n z^n} \times \frac{z^n - v^n}{v^n - z^n} = \frac{1}{v^n z^n} \times \frac{v^n - z^n}{v^n - z^n} \times \frac{v^n - z^n}{v^n - z^n} \times \frac{v^n - z^n}{v^n - z^n} \times \ldots, &c. \] \[ \frac{1}{v^n z^{n+1}} \times \frac{v^{m-1} + v^{m-2} z + v^{m-3} z^2 + \ldots}{v^{n-1} + v^{n-2} z + v^{n-3} z^2 + \ldots} = (\text{when } v = z) - \frac{mv^{m-1}}{v^{2m} \times nv^{n-1}} \] \[= \frac{-mv^{m-n}}{n}, \text{ which is the same as the expression } \left( \frac{mv^{m-n}}{n} \right) \text{ before derived with only the sign of } m \text{ changed. The remainder of the process being the same as before, shews that the series is general, or extends to all cases, regard being had to the signs. Q.E.D.}\]