Question

If \( \sum_{x=1}^{n^{2}} \sqrt{\left(x \cos \left(\cot ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right)^{2}-1}=A n^{4}+B n^{2} \), then \( 2(A+B) \) equals ________.

Hint:

With \( \theta=\cot^{-1}x \), \( \cos\theta=x/\sqrt{1+x^2} \) and \( \sin\theta=1/\sqrt{1+x^2} \); the radicand simplifies to \( x^2 \).

Answer 1

Correct Answer: 2

Let \( \theta=\cot^{-1}x \) with \( x\ge1 \). Then \( \cos\theta=\dfrac{x}{\sqrt{1+x^2}} \), \( \sin\theta=\dfrac{1}{\sqrt{1+x^2}} \).
\( x\cos\theta+\sin\theta=\dfrac{x^2+1}{\sqrt{1+x^2}}=\sqrt{1+x^2} \).
So \( \left(x\cos\theta+\sin\theta\right)^2-1=x^2 \) and the term is \( \sqrt{x^2}=x \).
The sum is \( \sum_{x=1}^{n^2}x=\dfrac{n^2(n^2+1)}{2}=\dfrac12 n^4+\dfrac12 n^2 \).
Hence \( A=B=\dfrac12 \) and \( 2(A+B)=2 \).
Answer: 2