Question

For any positive integer $n$, Let $f_n : [0, \infty) \rightarrow R$ as $f_n(x) = \sum_{i=1}^{n} \tan^{-1} \left( \frac{1}{1+(x+i)(x+i-1)} \right) \forall x \in [0, \infty)$ then find value of $\sum_{i=1}^{5} \tan^2(f_i(0))$.

Answer 1

55

Answer 2

The given function is $f_n(x) = \sum_{i=1}^{n} \tan^{-1} \left( \frac{1}{1+(x+i)(x+i-1)} \right)$.

We can simplify the term inside the $\tan^{-1}$ function. Let $y = x+i-1$. Then $x+i = y+1$. The term is $\tan^{-1} \left( \frac{1}{1+y(y+1)} \right)$.

Using the identity $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$, we can write $\frac{1}{1+y(y+1)}$ as $\frac{(y+1)-y}{1+y(y+1)}$.

So, $\tan^{-1} \left( \frac{1}{1+y(y+1)} \right) = \tan^{-1}(y+1) - \tan^{-1}(y)$.

Substituting back $y = x+i-1$, we get $\tan^{-1} \left( \frac{1}{1+(x+i)(x+i-1)} \right) = \tan^{-1}((x+i-1)+1) - \tan^{-1}(x+i-1) = \tan^{-1}(x+i) - \tan^{-1}(x+i-1)$.

Now, we can write $f_n(x)$ as a telescoping sum: $f_n(x) = \sum_{i=1}^{n} [\tan^{-1}(x+i) - \tan^{-1}(x+i-1)]$ $f_n(x) = (\tan^{-1}(x+1) - \tan^{-1}(x)) + (\tan^{-1}(x+2) - \tan^{-1}(x+1)) + \dots + (\tan^{-1}(x+n) - \tan^{-1}(x+n-1))$ All intermediate terms cancel out, leaving: $f_n(x) = \tan^{-1}(x+n) - \tan^{-1}(x)$.

We need to find the value of $\sum_{i=1}^{5} \tan^2(f_i(0))$. First, let's find $f_i(0)$. Substitute $x=0$ into the expression for $f_n(x)$: $f_n(0) = \tan^{-1}(0+n) - \tan^{-1}(0) = \tan^{-1}(n) - 0 = \tan^{-1}(n)$. So, $f_i(0) = \tan^{-1}(i)$ for $i=1, 2, 3, 4, 5$.

Next, we need to calculate $\tan^2(f_i(0))$: $\tan^2(f_i(0)) = \tan^2(\tan^{-1}(i))$. For any real number $k$, $\tan(\tan^{-1}(k)) = k$. So, $\tan^2(\tan^{-1}(i)) = (i)^2 = i^2$.

Finally, we need to calculate the sum $\sum_{i=1}^{5} \tan^2(f_i(0))$: $\sum_{i=1}^{5} \tan^2(f_i(0)) = \sum_{i=1}^{5} i^2$. This is the sum of the squares of the first 5 positive integers: $\sum_{i=1}^{5} i^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55$.

The value of the sum is 55.