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 $x + \frac{1}{x} = -1$.
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:
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:
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 $
Let f and g be functions satisfying $f(x+y) = f(x)f(y), f(1) = 7$ and $g(x+y) = g(xy), g(1) = 1$, for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.