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Let f(x0=a0+a1x+a2x^2++an x^n+ and (f(x)...

Let `f(x0=a_0+a_1x+a_2x^2++a_n x^n+` and `(f(x))/(1-x)=b_0+b_1x+b_2x^2++b_n x^n+` , then `b_n+b_(n-1)=a_n` b. `b_n-b_(n-1)=a_n` c. `b_n//b_(n-1)=a_n` d. none of these

A

`b_(n)+b_(n-1)= a_(n)`

B

`b_(n) - b_(n-1)= a_(n)`

C

`b_(n)//b_(n-1) = a_(n)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
`(f(x))/(1-x) = b_(0) + b_(1)x+b_(2)x^(2)+"…."+b_(n)x^(n)+"….."`
`rArr a_(0) + a_(1)x +a_(2)x^(2) + "….." + a_(n)x^(n) + "…."`
`= (1-x)(b_(0) + b_(1)x+b_(2)x^(2) + "….." + b_(n)x^(n)+ "…..")`
Comparing the coefficient of `x^(n)` on the both sides,
`a_(n) = b_(n) - b_(n-1)`.
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