Question

Given that \[ \sum_{k=1}^{n} k(k-1)=\frac{n(n-1)(n+1)}{3} \] and $\omega$ and $\omega^2$ are complex cube roots of unity. If \[ \sum_{k=1}^{2026} \left( k+\frac{1}{\omega} \right) \left( k+\frac{1}{\omega^2} \right) = \frac{2026}{3}(N+3), \] then $N=$

• A. $2025\times2026$
• B. $2025\times2024$
• C. $2027\times2025$
• D. $2026\times2027$

Hint:

For cube roots of unity: \[ 1+\omega+\omega^2=0 \] is the most important identity and is frequently used in simplification problems.

Answer 1

The Correct Option is C

Concept: Cube roots of unity satisfy: \[ 1+\omega+\omega^2=0 \] and \[ \omega^3=1 \] Also, \[ \frac{1}{\omega}=\omega^2, \qquad \frac{1}{\omega^2}=\omega \] These identities simplify algebraic expressions involving complex roots of unity.

Step 1: Simplify the given expression.
\[ \left( k+\frac{1}{\omega} \right) \left( k+\frac{1}{\omega^2} \right) = (k+\omega^2)(k+\omega) \] Expanding: \[ = k^2+k(\omega+\omega^2)+\omega^3 \] Using: \[ \omega+\omega^2=-1 \] and \[ \omega^3=1 \] we get: \[ = k^2-k+1 \] Therefore, \[ \sum_{k=1}^{2026} \left( k+\frac{1}{\omega} \right) \left( k+\frac{1}{\omega^2} \right) = \sum_{k=1}^{2026}(k^2-k+1) \]

Step 2: Separate the summation.
\[ = \sum_{k=1}^{2026}k(k-1) + \sum_{k=1}^{2026}1 \] Using the given identity: \[ \sum_{k=1}^{n}k(k-1) = \frac{n(n-1)(n+1)}{3} \] for \[ n=2026, \] we get: \[ = \frac{2026(2025)(2027)}{3} \] Also, \[ \sum_{k=1}^{2026}1=2026 \] Therefore, \[ = \frac{2026(2025)(2027)}{3}+2026 \]

Step 3: Take common factor.
\[ = \frac{2026}{3} \left[ 2025\times2027+3 \right] \] Comparing with: \[ \frac{2026}{3}(N+3) \] we obtain: \[ N=2025\times2027 \] Hence, \[ \boxed{ 2027\times2025 } \]