If \(a+\displaystyle\frac{1}{a}=-2\), then \(a^{17}+\displaystyle\frac{1}{a^{102}}\) is equal to:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
The condition fixes \(a\) completely: clearing the fraction gives a perfect square equal to zero, so \(a=-1\). Powers of \(-1\) are \(-1\) for odd exponents and \(1\) for even exponents.
Multiplying by \(a\): \(a^2+1=-2a\), so \(a^2+2a+1=0\)
\((a+1)^2=0\), so \(a=-1\)
\(a^{17}=(-1)^{17}=-1\) (odd power)
\(a^{102}=(-1)^{102}=1\) (even power), so \(\displaystyle\frac{1}{a^{102}}=1\)
\(a^{17}+\displaystyle\frac{1}{a^{102}}=-1+1=0\)