Question

If \(\sum _{n = 1}^{2026}tan^{-1}(\frac{1}{n^2+n+1}) = tan^{-1}(1-\frac{1}{x})\), where \(x\neq 0\), then \(x =\)

• A. \(2028\)
• B. \(2026\)
• C. \(1014\)
• D. \(1013\)

Hint:

Write each term as a difference of two inverse tangents.

Answer 1

The Correct Option is C

Step 1: Check a small case:
For $n$ up to $1$: $\tan^{-1}\frac13 = \tan^{-1}2 - \tan^{-1}1$, which is true since $\frac{2 - 1}{1 + 2} = \frac13$.

Step 2: General sum:
For $N$ terms the sum is $\tan^{-1}\frac{N}{N + 2}$, which equals $\tan^{-1}\left(1 - \frac{2}{N + 2}\right)$.
With $N = 2026$, $\frac1x = \frac{2}{2028}$, so $x = 1014$.

Final Answer:
$x = 1014$, option (C). \[ \boxed{1014} \]