Q43 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics

Let $f(n)=A(-2)^n+B(-3)^n \forall A, B \in \mathbf{R}$ and $n \in \mathbf{N}-\{1,2\}$. If $f(n)+a f(n-1)+b f(n-2)=0$, then $(a+b)(b-a)=$

  • A
    0
  • B
    5
  • C
    7
  • Correct answer
    D
    11

Solution

$\begin{aligned} & \text { We have, } f(n)=A(-2)^n+B(-3)^n \\ & \qquad \begin{array}{r} f(n)+a f(n-1)+b f(n-2)=0 \\ \therefore \quad A(-2)^n+B(-3)^n+a\left(A(-2)^{n-1}+B(-3)^{n-1}\right) \\ \\ \quad+b\left(A(-2)^{n-2}+B(-3)^{n-2}=0\right. \\ \Rightarrow A(-2)^{n-2}[4-2 a+b]+(-3)^{n-2} B[9-3 a+b]=0 \end{array} \end{aligned}$ $\therefore$ It is possible only $4-2 a+b=0 \text { and } 9-3 a+b=0$ Solving, we get $a=5, b=6$ $\therefore \quad(a+b)(b-a)=(5+6)(6-5)=11$

Hence, the correct answer is option: 'D'.

Relevant Questions