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 mode of reasoning, the Fermatian inference, or, as it is sometimes improperly termed, "mathematical induction."

As an example of this reasoning, Euler's demonstration of the binomial theorem for integral powers may be given. The theorem is that (x + y)^n, where n is a whole number, may be expanded into the sum of a series of terms of which the first is x^{n}y^o and each of the others is derived from the next preceding by diminishing the exponent of x by 1 and multiplying by that exponent and at the same time increasing the exponent of y by 1 and dividing by that increased exponent. Now, suppose this proposition to be true for a certain exponent, n = M, then it must also be true for n = M + 1. For let one of the terms in the expansion of (x + y)^M be written $$\mathrm{A}x^p y^q$$. Then, this term with the two following will be

$$\mathrm{A}{x^p}{y^q} + \mathrm{A}\frac{p}{q+1}x^{p-1} y^{q+1} + \mathrm{A}\frac{p}{q+1} \frac{p-1}{q+2}x^{p-2} y^{q+2}$$

Now, when (x + y)^M is multiplied by x + y to give (x + y)^{M + 1}, we multiply first by x and then by y instead of by x and add the two results. When we multiply by x, the second of the above three terms will be the only one giving a term involving $$x^p y^{q+1}$$ and the third will be the only one giving a term in x^{p - 1}y^{q + 2}; and when we multiply by y the first will be the only term giving a term in $$x^p y^{q+1}$$, and the second will be the only term giving a term in x^{p - 1}y^{q + 2}. Hence, adding like terms, we find that the coefficient of $${x^p}{y^{q+1}}$$ in the expansion of (x + y)^{M + 1} will be the sum of the coefficients of the first two of the above three terms, and that the coefficient of x^{p - 1}y^{q + 2} will be the sum of the coefficients of the last two terms. Hence, two successive terms in the expansion of (x + y)^{M + 1} will be