Question

If $f\left(n\right)=2.cosnx.\forall nϵN.thenf\left(1\right).f(n+1)-f\left(n\right)=$

• A. $f(n+3)$
• B. $f(n+2)$
• C. $f(n+1)2$
• D. None of these

Answer 1

Answer: (B)

$f(n+2)$

We have,

$f\left(1\right)=2\cos x$

$f(n+1)=2\cos (n+1)x$

$f\left(n\right)=2\cos nx$

Now,

$f\left(1\right)f(n+1)=2[\cos (n+2)x+\cos nx]$

Hence,

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

$2cos(n+2)x=f(n+2)$

Hence, this is the answser.