Question

Let $f(n,\theta )=\frac{2n+1}{(1+n^2|\cos \theta |)(1+(n+1{)}^2|\cos \theta |)},$ where $n\in N$ and $\theta \in R.$
If the minimum value of $\sum _{n=1}^{10}f(n,\theta )$ is $\frac{p}{q}$, where $p,q$ are co-prime, then the value of $(p+q)$ is

Answer 1

We have to minimise $\sum _{n=1}^{10}\frac{2n+1}{(1+n^2|\cos \theta |)(1+(n+1{)}^2|\cos \theta |)}$
Here, $n$ is varying. So, to minimise the expression, $|\cos \theta |$ should be maximum as it is in the denominator.
$\therefore |\cos \theta |=1$

Now, the expression becomes
$\sum _{n=1}^{10}\frac{2n+1}{(1+n^2)(1+(n+1{)}^2)}$

$=\sum _{n=1}^{10}\frac{(1+(n+1{)}^2)-(1+n^2)}{(1+n^2)(1+(n+1{)}^2)}$

$=\sum _{n=1}^{10}(\frac{1}{1+n^2}-\frac{1}{1+(n+1{)}^2})$

$=\frac{1}{1+1^2}-\frac{1}{1+{11}^2}$

$=\frac{60}{122}=\frac{30}{61}=\frac{p}{q}$
Thus, $p+q=91$