Page:Elementary algebra (1896).djvu/416

398 Rh Ex. 2. Find the general term of {3x^2+x-2}{(x - 2)^2(1-2x)} when expanded in a series of ascending powers of x.

By Ex. 4, Art. 493, we have

{3x^2+x-2}{(x - 2)^2(1-2x)} = - {}{3(1-2x)} - {}{3(x-2)} - {}{(x-2)^2} = - {1}{3(1-2x)} + {5}{3(2-x)} - {4}{(2-x)^2} = - {1}{3(1-2x)} + {5}{6(1-x 2)} - {4}{4(1-x 2)^2} = - {1}{3}(1-2x)^-1 + {5} {6}(1-x 2)^-1 - (1-x 2)^-2

Hence the (r + 1)th or general term of the expansion is

496. The following example sufficiently illustrates the method to be pursued when the denominator contains a quadratic factor.

Ex. Expand {7+x}{(1+x)(1+x^2) } in ascending powers of x and find the general term.

Assume {7+x}{(1+x)(1+x^2) } = {A}{1+x} + {B}{1+x^2}; 7 + x = A(1 + x^2) + (Bx+ C)( 1+ x).

Let 1 + x = 0, then A = 3;

equating the absolute terms, 7 = A+C, whence C = 4; equating the coefficients of x^2, = A+B, whence B = -3.

{7+x}{(1+x)(1+x^2) } = {3}{1+x} + {4-3x}{1+x^2} = 3(1 + x)^{-1} +(4 - 3x)(1+ x^2)^{-1} = 3{1 - x + x^2 + + (-1)^px^p +} + (4 -3x){1 - x^2 + x^4 + + (-1)^px^{2p} +}.