# 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. **Author(s):** Joseph Banks, William Sewell **Year:** 1796 **Journal:** Philosophical Transactions of the Royal Society of London **Volume:** 86 **Pages:** 4 pages **Identifier:** jstor-107011 **JSTOR URL:** --- 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 \[ \frac{1}{v^n z^n} \times \frac{z^n - v^n}{v^n - z^n} = \frac{1}{v^n z^n} \times \frac{v^n - z^n}{v^n - z^n} \times \frac{v^n - z^n}{v^n - z^n} \times \frac{v^n - z^n}{v^n - z^n} \times \ldots, &c. \] \[ \frac{1}{v^n z^{n+1}} \times \frac{v^{m-1} + v^{m-2} z + v^{m-3} z^2 + \ldots}{v^{n-1} + v^{n-2} z + v^{n-3} z^2 + \ldots} = (\text{when } v = z) - \frac{mv^{m-1}}{v^{2m} \times nv^{n-1}} \] \[= \frac{-mv^{m-n}}{n}, \text{ which is the same as the expression } \left( \frac{mv^{m-n}}{n} \right) \text{ before derived with only the sign of } m \text{ changed. The remainder of the process being the same as before, shews that the series is general, or extends to all cases, regard being had to the signs. Q.E.D.}\]