Question

If $f\left(n\right)=2\cos nx\forall n\in N,$ then $f\left(1\right)f(n+1)-f\left(n\right)$ is equal to

• A. $f(n+3)$
• B. $f(n+2)$
• C. $f(n+1)f\left(2\right)$
• D. $f(n+2)f\left(2\right)$

Answer 1

The correct option is B $f(n+2)$

$f\left(n\right)=2\cos nx$
$g=f\left(1\right)f(n+1)-f\left(n\right)$

$\Rightarrow g=4\cos x\cos (n+1)x-2\cos nx$

$\Rightarrow g=2[\cos (n+2)x+\cos nx]-2\cos nx\{\because 2\cos A\cos B=\cos (A+B)+\cos (A-B)\}$

$\Rightarrow g=2\cos (n+2)x$
$\therefore g=f(n+2)$
Ans: B