Question

If $x^2 + x + 1 = 0$, find the value of
$\sum_{k=1}^{15} \left(x^k + \frac{1}{x^k}\right)^4$

• A. 90
• B. 120
• C. 150
• D. 180

Hint:

Whenever you see the equation $x^2+x+1=0$ or $x^3-1=0$, immediately think of the cube roots of unity ($\omega$). Problems involving summations of powers of $\omega$ often rely on the periodic nature of these powers (repeating every 3 terms). Group the terms based on whether the power is a multiple of 3 or not.

Answer 1

The Correct Option is A

Step 1: Use nature of roots of the given equation

The equation
x² + x + 1 = 0

has roots which are the complex cube roots of unity other than 1. Let the roots be ω and ω², where:

ω³ = 1,   ω ≠ 1

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Step 2: Simplify the general term of the sum

Consider the expression:

(x^k + 1 / x^k)⁴

For x = ω or ω², since 1 / ω = ω², we get:

x^k + 1 / x^k = ω^k + ω^2k

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Step 3: Use identity of cube roots of unity

We know the identity:

1 + ω + ω² = 0

This implies:

ω^k + ω^2k =

• 2,  if k is a multiple of 3
• −1, if k is not a multiple of 3

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Step 4: Raise the expression to the fourth power

If k is a multiple of 3:

(ω^k + ω^2k)⁴ = 2⁴ = 16

If k is not a multiple of 3:

(−1)⁴ = 1

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Step 5: Count number of such terms from k = 1 to 15

Multiples of 3 between 1 and 15 are:
3, 6, 9, 12, 15 → 5 terms

Remaining terms:
15 − 5 = 10 terms

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Step 6: Calculate the total sum

Sum = (5 × 16) + (10 × 1)

Sum = 80 + 10

Sum = 90

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Final Answer:

The value of the given summation is
90