Page:The Monist Volume 2.djvu/552

For let one of the terms in the expansion of $(x+y)^{M}$ be written $Ax^{p}y^{q}$ Then, this term with the two following will be

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

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 and the second will be the only term giving a term in 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

$A\left(1+{\frac {p}{q+1}}\right)x^{p}y^{q+1}+A{\frac {p}{q+1}}\left(1+{\frac {p-1}{q+2}}\right)=A{\frac {p+q+1}{q+1}}x^{p}y^{q+1}+A{\frac {p+q+1}{q+1}}{\frac {p}{q+2}}x^{p-1}y^{q+2}$

It is, thus, seen that the succession of terms follows the rule. Thus if any integral power follows the rule, so also does the next higher power. But the first power obviously follows the rule. Hence, all powers do so.

Such reasoning holds good of any collection of objects capable of being ranged in a series which though it may be endless, can be numbered so that each member of it receives a definite integral number. For instance, all the whole numbers constitute such a numerable collection. Again, all numbers resulting from operating according to any definite rule with any finite number of whole numbers form such a collection. For they may be arranged in a series thus. Let F be the symbol of operation. First operate on 1, giving F(1). Then, operate on a second 1, giving F(1,1). Next, introduce 2, giving 3rd, F(2); 4th, F(2,1); 5th, F(1,2); 6th, F(2,2). Next use a third variable giving 7th, F(1,1,1); 8th, F(2,1,1); 9th, F(1,2,1); 10th, F(2,2,1); 11th, F(1,1,2); 12th, F(2,1,2); 13th, F(1,2,2);