Question

Let $f\left(x\right)=a_0+a_{1}x+a_2{x}^2+.....+a_n{x}^n+.......$ and $\frac{f\left(x\right)}{1-x}=b_0+b_{1}x+b_2{x}^2+.......+b_n{x}^n+.....$ then

• 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

Answer 1

The correct option is B $b_n-b_{n-1}=a_n$

$f\left(x\right)=a_0+a_{1}x+a_2{x}^2+.....+a_n{x}^n+.....$

$\frac{f\left(x\right)}{1-x}=b_0+b_{1}x+b_2{x}^2+....+b_n{x}^n+....$

$f\left(x\right)=(b_0+b_{1}x+b_2{x}^2+.....)-(b_{0}x+b_1{x}^2+......)$

$=b_0+(b_1-b_0)x+(b_2-b_1)x^2+.......$

$\Rightarrow a_n=b_n-b_{n-1}$