Mathematics · 2021
JEE Main · 3 August 2021, Shift 1 · Q81
If x^2-x+1=0, then the value of Σ_n=1^5(x^n+1/x^n)^2 is \ \ \ \.
If $\displaystyle x^2-x+1=0$, then the value of $\displaystyle \sum_{n=1}^5\left(x^n+\frac{1}{x^n}\right)^2$ is \_\_\_\_.
Official answer
8
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.