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Question

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

The correct answer is
175

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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Similar Questions

  1. Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
  2. Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
  3. Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
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Important Questions from Algebra

  1. Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
  2. Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
  3. Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
  4. If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
  5. Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
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