Question

The value of $\cot\left(\sum_{n=1}^{23}\cot^{-1}\left(1+\sum_{k=1}^{n}2k\right)\right)$ is

• A. $\frac{23}{25}$
• B. $\frac{25}{23}$
• C. $\frac{23}{24}$
• D. $\frac{24}{23}$

Hint:

Logic Tip: Telescoping series in inverse trigonometry almost always rely on algebraic manipulation of the numerator to match the factors of the denominator's second term. If the denominator is $1+A$, factor $A$ into $x \cdot y$ such that $x-y = \text{numerator}$.

Answer 1

The Correct Option is B

Concept:
Evaluate the innermost summation first, then use the property $\cot^{-1}(x) = \tan^{-1}(\frac{1}{x})$ for $x>0$. Convert the general term of the series into a difference of two inverse tangent functions to create a telescoping series, using the formula $\tan^{-1}\left(\frac{x-y}{1+xy}\right) = \tan^{-1}x - \tan^{-1}y$. 
Step 1: Evaluate the innermost sum and simplify the term. 
The inner sum is the sum of the first $n$ even numbers: $$\sum_{k=1}^{n}2k = 2\left(\frac{n(n+1)}{2}\right) = n(n+1)$$ Substitute this into the expression: $$\cot^{-1}\left(1+\sum_{k=1}^{n}2k\right) = \cot^{-1}(1+n(n+1))$$ Convert the inverse cotangent to inverse tangent: $$= \tan^{-1}\left(\frac{1}{1+n(n+1)}\right)$$ 
Step 2: Rewrite the term to form a telescoping series. 
We can rewrite the numerator $1$ as $(n+1) - n$: $$= \tan^{-1}\left(\frac{(n+1) - n}{1 + (n+1)n}\right)$$ Using the identity $\tan^{-1}\left(\frac{x-y}{1+xy}\right) = \tan^{-1}x - \tan^{-1}y$, where $x = n+1$ and $y = n$: $$= \tan^{-1}(n+1) - \tan^{-1}(n)$$ 
Step 3: Evaluate the summation from n=1 to 23. 
Now apply the summation $\sum_{n=1}^{23}$: $$S = \sum_{n=1}^{23} [\tan^{-1}(n+1) - \tan^{-1}(n)]$$ Write out the first few terms and the last term: $$S = (\tan^{-1}2 - \tan^{-1}1) + (\tan^{-1}3 - \tan^{-1}2) + ....... + (\tan^{-1}24 - \tan^{-1}23)$$ Since it is a telescoping series, all intermediate terms cancel out, leaving only the first and last parts: $$S = \tan^{-1}24 - \tan^{-1}1$$ 
Step 4: Combine the resulting inverse tangents and evaluate the final expression. 
Use the identity $\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$ again: $$S = \tan^{-1}\left(\frac{24 - 1}{1 + 24(1)}\right) = \tan^{-1}\left(\frac{23}{25}\right)$$ We need to find the cotangent of this sum: $$\cot(S) = \cot\left(\tan^{-1}\left(\frac{23}{25}\right)\right)$$ Using the property $\tan^{-1}(x) = \cot^{-1}(1/x)$: $$\cot(S) = \cot\left(\cot^{-1}\left(\frac{25}{23}\right)\right) = \frac{25}{23}$$