If $x^2-x+1=0$, then $\sum_\limits {n=1}^5\left(x^n+\frac{1}{x^n}\right)=$

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

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Application +4 | -1

x^2-x+1=0

, then

\sum\limits_{{n=1}}^5\left(x^n+\frac{1}{x^n}\right)=

View answer and explanation

Correct option: A

Option A (8) is correct because it represents the sum of the given expression for the roots of the quadratic equation.

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\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2

a

d

\left( {n + 1} \right)

a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ...

(1-x)^{201}\left(1+x+x^2\right)^{200}

\left(\frac{1}{x}+2 x\right)^n

{x^5}

\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1,

\left(\begin{array}{c}2 n+1 \\ 0\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 6\end{array}\right)+\ldots=170

n