tan 
−1
 ( 
1+r+r 
2
 
1
​	
 )=tan 
−1
 ( 
1+(r+1)r
(r+1)−r
​	
 ) 
=tan 
−1
 (r+1)−tan 
−1
 r 

So ∑ 
r=1
n
​	
 tan 
−1
 ( 
1+r+r 
2
 
1
​	
 )=(tan 
−1
 2−tan 
−1
 1)+(tan 
−1
 3−tan 
−1
 2)+………+(tan 
−1
 (n+1)− tan 
−1
 n) 

⇒
​	
  
=tan 
−1
 (n+1)−tan 
−1
 1
n→∞
lim
​	
 tan{ 
r=1
∑
n
​	
 tan 
−1
 ( 
1+r+r 
2
 
1
​	
 )}
n→∞
lim
​	
 tan(tan 
−1
 (n+1)−tan 
−1
 1)
=tan( 
2
π
​	
 − 
4
π
​	
 )=1
​	
 

Let's break down and analyze the problem step by step to arrive at the solution clearly.

### Step 1: Understanding the expression

We are given:
\[
\tan^{-1}\left(\frac{1 + r + r^2}{1}\right) = \tan^{-1}\left(\frac{1 + (r+1)r}{(r+1)-r}\right) = \tan^{-1}(r+1) - \tan^{-1}(r)
\]
This shows that the given expression can be simplified to the difference of two arctangents:
\[
\tan^{-1}(r+1) - \tan^{-1}(r)
\]

### Step 2: Summation

The next step is to apply this to the summation:
\[
\sum_{r=1}^{n} \tan^{-1}\left(\frac{1 + r + r^2}{1}\right)
\]
This becomes:
\[
\sum_{r=1}^{n} \left(\tan^{-1}(r+1) - \tan^{-1}(r)\right)
\]
This is a **telescoping series**, where most terms cancel out. Expanding the series:
\[
\left(\tan^{-1}(2) - \tan^{-1}(1)\right) + \left(\tan^{-1}(3) - \tan^{-1}(2)\right) + \cdots + \left(\tan^{-1}(n+1) - \tan^{-1}(n)\right)
\]
All the intermediate terms cancel, leaving:
\[
\tan^{-1}(n+1) - \tan^{-1}(1)
\]

### Step 3: Taking the limit as \(n \to \infty\)

Now, consider the limit as \(n \to \infty\):
\[
\lim_{n \to \infty} \left(\tan^{-1}(n+1) - \tan^{-1}(1)\right)
\]
As \(n \to \infty\), we know that:
\[
\tan^{-1}(n+1) \to \frac{\pi}{2}
\]
Thus:
\[
\lim_{n \to \infty} \left(\frac{\pi}{2} - \tan^{-1}(1)\right) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}
\]

### Step 4: Final expression

The final part of the problem is:
\[
\lim_{n \to \infty} \tan\left(\tan^{-1}(n+1) - \tan^{-1}(1)\right)
\]
We already calculated that:
\[
\tan^{-1}(n+1) - \tan^{-1}(1) \to \frac{\pi}{4}
\]
Thus:
\[
\tan\left(\frac{\pi}{4}\right) = 1
\]

### Conclusion

Therefore, the result of the given expression is:
\[
\boxed{1}
\]

---

Do you need more details or have any questions? Here are some related questions you might want to explore:

1. What is the behavior of the arctangent function as \(r \to \infty\)?
2. How can we identify and simplify a telescoping series?
3. How does the limit of arctangent functions behave?
4. Why does \(\tan\left(\frac{\pi}{4}\right) = 1\)?
5. How can we generalize this result for other similar functions?

**Tip:** Telescoping series are very useful in simplifying large sums, especially when terms cancel out step by step.

tan^{-1} \left( \frac{1 + r + r^2}{1} \right) = tan^{-1} \left( \frac{1 + (r+1)r}{(r+1)-r} \right) = tan^{-1}(r+1) - tan^{-1}(r) So ∑_{r=1}^{n} tan^{-1} \left( \frac{1 + r + r^2}{1} \right) = (tan^{-1} 2 − tan^{-1} 1) + (tan^{-1} 3 − tan^{-1} 2) +……… + (tan^{-1}(n+1) − tan^{-1} n) ⇒ = tan^{-1}(n+1) − tan^{-1} 1 n→∞lim tan{ ∑_{r=1}^{n} tan^{-1} \left( \frac{1 + r + r^2}{1} \right) } n→∞lim tan(tan^{-1}(n+1) − tan^{-1} 1) = tan( \frac{2}{π} − \frac{4}{π} ) = 1

Explore the solution to a trigonometric problem involving the summation of arctangent terms. The problem is solved using the concept of telescoping series, with a step-by-step explanation of how to apply arctangent limits as n approaches infinity.

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