Equation Analysis
We are given the equation $x^2 + x + 1 = 0$. Multiplying by $(x-1)$ gives:
$ (x-1)(x^2 + x + 1) = 0 $
$ x^3 - 1 = 0 $
$ x^3 = 1 $
This indicates that $x$ is a complex cube root of unity. Additionally, dividing the original equation $x^2 + x + 1 = 0$ by $x$ (note $x \neq 0$), we get $x + 1 + \frac{1}{x} = 0$, which simplifies to $x + \frac{1}{x} = -1$.
Term Evaluation
The expression involves terms of the form $\left(x^n + \frac{1}{x^n}\right)^4$. Let's evaluate this based on $n$ modulo 3:
- If $n \pmod 3 = 1$:
$x^n + \frac{1}{x^n} = x^1 + \frac{1}{x^1} = x + \frac{1}{x} = -1$.
The term value is $(-1)^4 = 1$. - If $n \pmod 3 = 2$:
$x^n + \frac{1}{x^n} = x^2 + \frac{1}{x^2}$. We calculate this using $(x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}$. Substituting $x + \frac{1}{x} = -1$, we get $(-1)^2 = x^2 + 2 + \frac{1}{x^2}$, which yields $x^2 + \frac{1}{x^2} = 1 - 2 = -1$.
The term value is $(-1)^4 = 1$. - If $n \pmod 3 = 0$:
$x^n + \frac{1}{x^n} = x^3 + \frac{1}{x^3} = 1 + \frac{1}{1} = 2$.
The term value is $(2)^4 = 16$.
Summation Calculation
We need to calculate the sum $S = \sum_{n=1}^{25} \left(x^n + \frac{1}{x^n}\right)^4$. There are 25 terms.
Let's count how many terms fall into each modulo case:
- Terms where $n \equiv 0 \pmod 3$ (multiples of 3): $3, 6, ..., 24$. There are $24 / 3 = 8$ such terms.
- Terms where $n \equiv 1 \pmod 3$: $1, 4, ..., 25$. There are $9$ such terms.
- Terms where $n \equiv 2 \pmod 3$: $2, 5, ..., 23$. There are $8$ such terms.
- Total terms checked: $8 + 9 + 8 = 25$.
The total sum is calculated by summing the values based on these counts:
Sum = (Number of terms for $n \pmod 3 \neq 0$) $\times$ (Value for $n \pmod 3 \neq 0$) + (Number of terms for $n \pmod 3 = 0$) $\times$ (Value for $n \pmod 3 = 0$)
Sum = $ ( (9 + 8) \times 1 ) + ( 8 \times 16 )$
Sum = $ (17 \times 1) + (8 \times 16)$
Sum = $ 17 + 128 $
Sum = $ 145 $
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Important Questions from Algebra
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