Q55 / 150 · MHT CET 2026 (15 April Shift 1) · Mathematics
If $\omega$ is a complex cube root of unity, then the value of the expression $2\left(1+\dfrac{1}{\omega}\right)\left(1+\dfrac{1}{\omega^2}\right) + 3\left(2+\dfrac{1}{\omega}\right)\left(2+\dfrac{1}{\omega^2}\right) + \ldots + (n+1)\left(n+\dfrac{1}{\omega}\right)\left(n+\dfrac{1}{\omega^2}\right)$ is...
- Correct answerA$\left[\dfrac{n(n+1)}{2}\right]^2 + n$
- B$\left[\dfrac{n(n+1)}{2}\right]^2 - n$
- C$\left[\dfrac{n(n+1)}{2}\right]^2$
- D$\left[\dfrac{n(n-1)}{2}\right]^2$
Solution
The general term of the given series is $T_k = (k+1)\left(k+\dfrac{1}{\omega}\right)\left(k+\dfrac{1}{\omega^2}\right)$.
Since $\omega^3 = 1$, we have $\dfrac{1}{\omega} = \omega^2$ and $\dfrac{1}{\omega^2} = \omega$.
$T_k = (k+1)(k+\omega^2)(k+\omega)$
$T_k = (k+1)(k^2 + k(\omega+\omega^2) + \omega^3)$
Using $1+\omega+\omega^2 = 0 \Rightarrow \omega+\omega^2 = -1$ and $\omega^3 = 1$, we get:
$T_k = (k+1)(k^2 - k + 1)$
$T_k = k^3 + 1$
The sum of the series up to $n$ terms is:
$S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (k^3 + 1)$
$S_n = \sum_{k=1}^n k^3 + \sum_{k=1}^n 1$
$S_n = \left[\dfrac{n(n+1)}{2}\right]^2 + n$
Since $\omega^3 = 1$, we have $\dfrac{1}{\omega} = \omega^2$ and $\dfrac{1}{\omega^2} = \omega$.
$T_k = (k+1)(k+\omega^2)(k+\omega)$
$T_k = (k+1)(k^2 + k(\omega+\omega^2) + \omega^3)$
Using $1+\omega+\omega^2 = 0 \Rightarrow \omega+\omega^2 = -1$ and $\omega^3 = 1$, we get:
$T_k = (k+1)(k^2 - k + 1)$
$T_k = k^3 + 1$
The sum of the series up to $n$ terms is:
$S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (k^3 + 1)$
$S_n = \sum_{k=1}^n k^3 + \sum_{k=1}^n 1$
$S_n = \left[\dfrac{n(n+1)}{2}\right]^2 + n$
Hence, the correct answer is option: 'A'.