If 1, omega, omega^2 are the cube roots of un | vedclass

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(B) The general term of the series is $T_n = n(n+1 + \frac{1}{\omega})(n+1 + \frac{1}{\omega^2})$.Since $\omega^3 = 1$,we have $\frac{1}{\omega} = \om

Vedclass · Accepted Answer

$3465$

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(B) The general term of the series is $T_n = n(n+1 + \frac{1}{\omega})(n+1 + \frac{1}{\omega^2})$.Since $\omega^3 = 1$,we have $\frac{1}{\omega} = \omega^2$ and $\frac{1}{\omega^2} = \omega$.Thus,$T_n = n(n+1 + \omega^2)(n+1 + \omega) = n((n+1)^2 + (n+1)(\omega + \omega^2) + \omega^3)$.Using the property $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.So,$T_n = n((n+1)^2 - (n+1) + 1) = n(n^2 + 2n + 1 - n - 1 + 1) = n(n^2 + n + 1) = n^3 + n^2 + n$.The sum of $10$ terms is $\sum_{n=1}^{10} (n^3 + n^2 + n) = \sum n^3 + \sum n^2 + \sum n$.Using standard summation formulas:$\sum_{n=1}^{10} n^3 = (\frac{10 	imes 11}{2})^2 = 55^2 = 3025$.$\sum_{n=1}^{10} n^2 = \frac{10 	imes 11 	imes 21}{6} = 385$.$\sum_{n=1}^{10} n = \frac{10 	imes 11}{2} = 55$.Total sum $= 3025 + 385 + 55 = 3465$.

If $1, \omega, \omega^2$ are the cube roots of unity,then
$1(2+\frac{1}{\omega})(2+\frac{1}{\omega^2})+2(3+\frac{1}{\omega})(3+\frac{1}{\omega^2})+3(4+\frac{1}{\omega})(4+\frac{1}{\omega^2})+\ldots 10 \text{ terms} =$

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If $1, \omega, \omega^2$ are the cube roots of unity,then$1(2+\frac{1}{\omega})(2+\frac{1}{\omega^2})+2(3+\frac{1}{\omega})(3+\frac{1}{\omega^2})+3(4+\frac{1}{\omega})(4+\frac{1}{\omega^2})+\ldots 10 \text{ terms} =$