Question

If f(x)=1/((1+2x)(3-x))
A)Express f(x) in partial fractions.
B)find the first four terms in the series expansion of f(x) in ascending powers of x.
C) find the coefficient of x^n in the series expansion of f x in ascending powers of x.

Answer 1

let 1/((1+2x)(3-x)) = A/(1+2x) + B(3-x) = ( A(3-x) + B(1+2x))/((1+2x)(3-x))

1 = A(3-x) + B(1+2x)
let x = 3 ---> 1 = 7B, or B = 1/7
let x= -1/2 ---> 1 = 7A/2 or A = 2/7

then 1/((1+2x)(3-x)) = 2/(7(1 + 2x)) + 1/(7(3 - x))

2/(7 + 14x) = (2/7)(1 + 2x)^-1
= (2/7)(1 + (-1)(2x) + (-1)(-2)/2!(2x)^2 + (-1)(-2)(-3)/3!(2x)^3 + .... )
= 2/7 - 4x/7 + 8x^2 /7 - 16x^3 / 7 + ....

1/(7(3 - x)) = (1/7)(3 - x)^-1
= (1/7)(3^-1 + (-1)(3^-2)(-x) + (-1)(-2)/2! (3^-3)(-x)^2 + (-1)(-2)(-3)/3! (3^-4)(-x)^3 + ...
= (1/7)(1/3 + x/9 + x^2/27 + x^3 / 81 + ...
= 1/21 + x/63 + x^2 /189 + x^3 / 567 + ...

then f(x) = 1/((1+2x)(3-x))
= 2/7 - 4x/7 + 8x^2 /7 - 16x^3 / 7 + .... + 1/21 + x/63 + x^2 /189 + x^3 / 567 + ...

= (2/7 + 1/21) + x(1/63 - 4/7) + x^2(1/189 + 8/7) + x^3(1/567 - 16/7) + ...
= 1/3 - (5/9)x + (31/27)x^2 + (-185/81)x^3 + ....

general term for (2/7)(1 + 2x)^-1
= (2/7)(-1)(-2)(-3)...(-n)/n! (1^(-1-n)(2x)^n
= +(2/7)(2^n)x^n , for even n, and -(2/7)(2^n)x^n for odd n

Ok, your turn to find the general term for (1/7)(3 - x)^-1

then add up the 2 general terms