1.

If the expansion in power of x of the function `(1)/(( 1 - ax)(1 - bx)) is a_(0) + a_(1) x + a_(2) x^(2) + a_(3) a^(3) + …, ` then `a_(n)` isA. `(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» Correct Answer - d
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) `.


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