If the expansion in powers of x be the function `1//[(1-ax)(1-bx)]` is `a_(0)+a_(1)x+a_(2)x^(2)+a_(3)x^(3)+"…."`, then `a_(n)` is

Options

A. `(b^(n) - a^(n))/(b-a)`
B. `(a^(n) - b^(n))/(b-a)`
C. `(a^(n+1) - b^(n+1))/(b-a)`
D. `(b^(n+1) - a^(n+1))/(b-a)`

Answer

d

Explanation

We have,
`(1)/((1 - ax)(1 - bx) )=(1)/(a -b) {(a)/(1 - ax) (b)/(1 - bx)}`
`rArr (1)/((1 - ax)(1 - bx) )=(1)/(a -b) {a(1 - ax)^(-1) - b(1 - bx)^(-1)}` ltbegt `rArr (1)/((1 - ax)(1 - bx) )=(1)/(a -b) {asum_(r=0)^(infty) ( ax)^(1) - b(1 - bx)^(-1)}`
`rArr (1)/((1 - ax)(1 - bx) )=(1)/(a -b) {a(sum_(r=0)^(infty) a^(r)x^(r)) - b(sum_(r= 0)^(infty) b^(r) x^(r))^()}`
`therefore a_(n)` = Coefficient of `x^(n)` in` (1)/((1 - ax) (1 - bx))`
`rArr a_(n) = (1_)/(a - b) { axx a^(n) - b xxb^(n)} = ((a^(n +1) - b^(n +1))/(a -b) `.