Question

In $u_n=2\cos n\theta$ and $u_1{u}_n-u_{n-1}$ is equal to

• A. $u_{n-2}$
• B. $u_{n+1}$
• C. $0$
• D. None of these

Answer 1

Answer: (B)

$u_{n+1}$

$u_n=2\cos n\theta$
$u_1{u}_n-u_{n-1}=2\cos \theta .2\cos n\theta -2\cos (n-1)\theta$
$=2\left(2\cos \theta \cos n\theta \right)-2\cos (n-1)\theta$
$2\cos A\cos B=\cos (A+B)+\cos (A-B)$
$u_1{u}_n-u_{n-1}=2[\cos (\theta +n\theta )+\cos (n\theta -\theta )]-2\cos (n-1)\theta$
$=2\cos (n+1)\theta +2\cos (n-1)\theta -2\cos (n-1)\theta$
$=2\cos (n+1)\theta$
$=u_{n+1}$