Question

If $\omega $ is a complex cube root of unity and $f(n)=2(1+\omega)\left(1+\omega^{2}\right)+3(2+\omega)\left(2+\omega^{2}\right)+\ldots(n+1)(n+\omega)\left(n+\omega^{2}\right)$, then the value of $f(19)$ is

• A. $36150$
• B. $36100$
• C. $36101$
• D. $36119$

Answer 1

Answer: (D)

We have,
$ \begin{array}{l} f(n)=2(1+\omega)\left(1+\omega^{2}\right)+3(2+\omega)\left(2+\omega^{2}\right)+\ldots(n+1)(n+\omega)\left(n+\omega^{2}\right) \\ \Rightarrow f(n)=\sum(n+1)(n+\omega)\left(n+\omega^{2}\right) \\ \Rightarrow f(n)=\sum(n+1)\left(n^{2}+n \omega^{2}+n \omega+\omega^{3}\right) \\ \Rightarrow f(n)=\sum(n+1)\left(n^{2}+n\left(\omega^{2}+\omega\right)+1\right) \\ \Rightarrow f(n)=\sum(n+1)\left(n^{2}-n+1\right) \\ \Rightarrow f(n)=\sum\left(n^{3}+1\right) \\ \Rightarrow f(n)=\left[\frac{n(n+1)}{2}\right]^{2}+n \\ \therefore f(19)=[190]^{2}+19=36119 \end{array} $