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: Understanding the Concept:
Use \(\tan^{-1}a - \tan^{-1}b = \tan^{-1}\frac{a - b}{1 + ab}\). Take \(a = n + 1\) and \(b = n\).

Step 2: Key Formula or Approach:
\(\tan^{-1}\frac{(n + 1) - n}{1 + n(n + 1)} = \tan^{-1}\frac{1}{n^2 + n + 1}\), so each term equals \(\tan^{-1}(n + 1) - \tan^{-1}n\).

Step 3: Detailed Explanation:
The sum telescopes: \(\sum_{n=1}^{2026}[\tan^{-1}(n+1) - \tan^{-1}n] = \tan^{-1}2027 - \tan^{-1}1\).
\[ = \tan^{-1}\frac{2027 - 1}{1 + 2027} = \tan^{-1}\frac{2026}{2028} = \tan^{-1}\frac{1013}{1014} \]
Set \(1 - \frac1x = \frac{1013}{1014}\), so \(\frac1x = \frac1{1014}\).
\[ x = 1014 \]

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