If the expansion in powers of x of the function (1/(1-ax)(1-bx)) is a0+a1x+a2x2+a3x3+ ..... , then an is :

Question

If the expansion in powers of x of the function (1/(1-ax)(1-bx)) is a0+a1x+a2x2+a3x3+ ..... , then an is : (A) (an-bn/b-a) (B) (an+1-bn+1/b-a) (C) (bn+1-an+1/b-a) (D) (bn-an/b-a)

Tardigrade Admin · Accepted Answer

∵(1-ax)-1(1-bx)-1 =(1+ax+a2x2+...)(1+bx+b2x2+...) Hence an = coefficient of xn in (1-ax)-1(1-bx)-1 =a0bn+abn-1+...+anb0 =a0bn((((a/b))n+1-1/(a)b-1))R =(bn(an+1-bn+1)/a-b)⋅(b/bn+1)=(an+1-bn+1/a-b)

Q. If the expansion in powers of $x$ of the function $\frac{1}{( 1 - a x ) ( 1 - b x )}$ is
$a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + .....,$ then $a_{n}$ is :

$\frac{a ^{n} - b ^{n}}{b - a}$

$\frac{a ^{n + 1} - b ^{n + 1}}{b - a}$

$\frac{b ^{n + 1} - a ^{n + 1}}{b - a}$

$\frac{b ^{n} - a ^{n}}{b - a}$

Q. If the expansion in powers of x of the function (1−ax)(1−bx)1​ is a0​+a1​x+a2​x2+a3​x3+....., then an​ is :

b−aan−bn​

b−aan+1−bn+1​

b−abn+1−an+1​

b−abn−an​

∵(1−ax)−1(1−bx)−1 =(1+ax+a2x2+...)(1+bx+b2x2+...) Hence an​= coefficient of xn in (1−ax)−1(1−bx)−1 =a0bn+abn−1+...+anb0 =a0bn(ba​−1(ba​)n+1−1​)R =a−bbn(an+1−bn+1)​⋅bn+1b​=a−ban+1−bn+1​

Q. If the expansion in powers of $x$ of the function $\frac{1}{( 1 - a x ) ( 1 - b x )}$ is
$a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + .....,$ then $a_{n}$ is :

$\frac{a ^{n} - b ^{n}}{b - a}$

$\frac{a ^{n + 1} - b ^{n + 1}}{b - a}$

$\frac{b ^{n + 1} - a ^{n + 1}}{b - a}$

$\frac{b ^{n} - a ^{n}}{b - a}$