If x^2 + x + 1 = 0$x^{2} + x + 1 = 0$ , then the value of left(mathrmx+frac1mathrmxright)^4+left(mathrmx^2+frac1mathrmx^2right)^4+left(mathrmx^3+frac1mathrmx^3right)^4+ldots+left(mathrmx^25+frac1mathrmx^25right)^4$\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{4}+\left(\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}\right)^{4}+\left(\mathrm{x}^{3}+\frac{1}{\mathrm{x}^{3}}\right)^{4}+\ldots+\left(\mathrm{x}^{25}+\frac{1}{\mathrm{x}^{25}}\right)^{4}$ is :

### Related Formula x^2 + x + 1 = 0 Rightarrow x = omega, omega^2$$x^2 + x + 1 = 0 \Rightarrow x = \omega, \omega^2$$ Properties of cube roots of unity: omega^3 = 1$\omega^3 = 1$ and 1 + omega + omega^2 = 0$1 + \omega + \omega^2 = 0$. ### Core Logic Let alpha = omega$\alpha = \omega$. Then frac1x = frac1omega = omega^2$\frac{1}{x} = \frac{1}{\omega} = \omega^2$. The series is sum_k=1^25 (omega^k + omega^2k)^4$\sum_{k=1}^{25} (\omega^k + \omega^{2k})^4$. Evaluate the term T_k = (omega^k + omega^2k)^4$T_k = (\omega^k + \omega^{2k})^4$ based on the modulo of k$k$ with 3. ### Step 1: Cyclic Evaluation Case 1: k = 3m$k = 3m$ (multiples of 3) T_3m = (omega^3m + omega^6m)^4 = (1 + 1)^4 = 2^4 = 16$$T_{3m} = (\omega^{3m} + \omega^{6m})^4 = (1 + 1)^4 = 2^4 = 16$$ There are 8$8$ such multiples up to 25 (3, 6, 9, dots, 24$3, 6, 9, \dots, 24$). Case 2: k neq 3m$k \neq 3m$ (non-multiples of 3) For k = 1, 2, 4, 5, dots$k = 1, 2, 4, 5, \dots$ omega^k + omega^2k$\omega^k + \omega^{2k}$ will always be omega + omega^2$\omega + \omega^2$ or omega^2 + omega$\omega^2 + \omega$. Since 1 + omega + omega^2 = 0$1 + \omega + \omega^2 = 0$, we have omega + omega^2 = -1$\omega + \omega^2 = -1$. T_k neq 3m = (-1)^4 = 1$$T_{k \neq 3m} = (-1)^4 = 1$$ There are 25 - 8 = 17$25 - 8 = 17$ such non-multiples up to 25. ### Step 2: Total Sum textSum = 17 times 1 + 8 times 16$$\text{Sum} = 17 \times 1 + 8 \times 16$$ = 17 + 128 = 145$$= 17 + 128 = 145$$ ### Pattern Recognition Powers of x+1/x$x+1/x$ when x$x$ solves x^2 pm x + 1 = 0$x^2 \pm x + 1 = 0$ perfectly orbit around periods of 3 or 6. Isolate the 3m$3m$ resonant beats (which hit pure scalars like 1+1=2$1+1=2$) versus the out-of-phase beats (which collapse to -1$-1$ or 1$1$ via basic omega$\omega$ identities). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations