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Thus the series is a recurring series in which the scale of relation is .

516. To find any term when the scale of relation is given. If the scale of relation of a recurring series is given, any term can be found when a sufficient number of the preceding terms are known. As the method of procedure is the same, however many terms the scale of relation may consist of, the following illustration will be sufficient:

If $1-px-qx^{2}-rx^{3}$

is the scale of relation of the series we have , or ; thus any coefficient can be found when the coefficients of the three preceding terms are known.

517. To find the scale of relation. If a sufficient number of the terms of a series be given, the scale of relation may be found.

Ex. Find the scale of relation of the recurring series $2+5x+13x^{2}+35x^{3}+97x^{4}+275x^{5}+798x^{6}+\ldots$ .

This is plainly not a series of the first order. If it be of the second order, to obtain p and g we have the equations a

$13=5p+2q$, and $35=13p4+5q$;

whence p = 5, and q=-6. By using these values of p and g, we can obtain the fifth and sixth coefficients ; hence they are correct, and the scale of relation is

$1-5x+6x^{2}$.

If we could not have obtained the remaining coefficients with these values of p and g, we would have assumed the series to be of the third order, and formed the equations

$35=13p+5q+2r$, $97=35p+13q+5r$, $275=97p+35q+187$;