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(C) Given,$\sum_{k=1}^n\left(k+\frac{1}{\omega}\right)\left(k+\frac{1}{\omega^2}\right)=340$ Expanding the term inside the summation: $\sum_{k=1}^n\le

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) Given,$\sum_{k=1}^n\left(k+\frac{1}{\omega}\right)\left(k+\frac{1}{\omega^2}\right)=340$ Expanding the term inside the summation: $\sum_{k=1}^n\left(k^2+k\left(\frac{1}{\omega^2}+\frac{1}{\omega}\right)+\frac{1}{\omega^3}\right)=340$ Since $\omega^3=1$,we have $\frac{1}{\omega}=\omega^2$ and $\frac{1}{\omega^2}=\omega$. Also,$1+\omega+\omega^2=0 \implies \omega+\omega^2=-1$. Substituting these values: $\sum_{k=1}^n\left(k^2+k(-1)+1\right)=340$ $\sum_{k=1}^n(k^2-k+1)=340$ Using summation formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$: $\frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} + n = 340$ $\frac{n(n+1)}{2} \left[ \frac{2n+1}{3} - 1 \right] + n = 340$ $\frac{n(n+1)}{2} \left[ \frac{2n-2}{3} \right] + n = 340$ $\frac{n(n+1)(n-1)}{3} + n = 340$ $\frac{n(n^2-1)}{3} + n = 340$ $n^3-n+3n = 1020$ $n^3+2n = 1020$ Testing values,for $n=10$: $10^3 + 2(10) = 1000 + 20 = 1020$. Thus,$n=10$.

If $\omega$ represents a cube root of unity and $\sum_{k=1}^n\left(k+\frac{1}{\omega}\right)\left(k+\frac{1}{\omega^2}\right)=340$,then $n=$

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