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Question: Let \(\omega\) is an imaginary cube root of unity then the value of \(2(\omega + 1)(\omega^{2} + 1)...

Let ω\omega is an imaginary cube root of unity then the value of

2(ω+1)(ω2+1)+3(2ω+1)(2ω2+1)+.....+(n+1)(nω+1)(nω2+1)2(\omega + 1)(\omega^{2} + 1) + 3(2\omega + 1)(2\omega^{2} + 1) + ..... + (n + 1)(n\omega + 1)(n\omega^{2} + 1) is

A

[n(n+1)2]2+n\left\lbrack \frac{n(n + 1)}{2} \right\rbrack^{2} + n

B

[n(n+1)2]2\left\lbrack \frac{n(n + 1)}{2} \right\rbrack^{2}

C

[n(n+1)2]2−n\left\lbrack \frac{n(n + 1)}{2} \right\rbrack^{2} - n

D

None of these

Answer

[n(n+1)2]2+n\left\lbrack \frac{n(n + 1)}{2} \right\rbrack^{2} + n

Explanation

Solution

Sol. 2(ω+1)(ω2+1)+3(2ω+1)(2ω2+1)+.....+(n+1)(nω+1)(nω2+1)=∑r=1n(r+1)(rω+1)(rω2+1)2(\omega + 1)(\omega^{2} + 1) + 3(2\omega + 1)(2\omega^{2} + 1) + ..... + (n + 1)(n\omega + 1)(n\omega^{2} + 1) = \sum_{r = 1}^{n}{(r + 1)(r\omega + 1})(r\omega^{2} + 1)

=∑r=1n(r+1)(r2ω3+rω+rω2+1)=∑r=1n(r+1)(r2−r+1)=∑r=1n(r3−r2+r+r2−r+1)\mathbf{=}\sum_{\mathbf{r = 1}}^{\mathbf{n}}{\mathbf{(r + 1)(}\mathbf{r}^{\mathbf{2}}\mathbf{\omega}^{\mathbf{3}}\mathbf{+ r\omega +}}\mathbf{r}\mathbf{\omega}^{\mathbf{2}}\mathbf{+ 1) =}\sum_{\mathbf{r = 1}}^{\mathbf{n}}{\mathbf{(r + 1)(}\mathbf{r}^{\mathbf{2}}\mathbf{- r + 1)}}\mathbf{=}\sum_{\mathbf{r = 1}}^{\mathbf{n}}{\mathbf{(}\mathbf{r}^{\mathbf{3}}\mathbf{-}\mathbf{r}^{\mathbf{2}}\mathbf{+ r +}\mathbf{r}^{\mathbf{2}}\mathbf{- r + 1)}} =∑r=1n(r3)+∑r=1n(1)=[n(n+1)2]2+n.\mathbf{=}\sum_{\mathbf{r = 1}}^{\mathbf{n}}{\mathbf{(}\mathbf{r}^{\mathbf{3}}\mathbf{)}}\mathbf{+}\sum_{\mathbf{r = 1}}^{\mathbf{n}}\mathbf{(1)}\mathbf{=}\left\lbrack \frac{\mathbf{n(n + 1)}}{\mathbf{2}} \right\rbrack^{\mathbf{2}}\mathbf{+ n.}