Question

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

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

Answer 1

The correct option is B $u_{n+1}$

We have, $u_1{u}_n-u_{n-1}$

$=2\cos \left(\theta \right).2\cos \left(n\theta \right)-2\cos (n-1)\theta$

$=2\left[2\cos \right(\theta ).\cos (n\theta )-\cos (n-1\left)\theta \right]$

$=2\left[\cos \right(n+1)\theta +\cos (n-1)\theta -\cos (n-1\left)\theta \right]$

$=2\cos (n+1)\theta$

$=u_{n+1}$.