Complexity! (7)

Algebra Level 4

Evaluate the following sum in terms of $n$ ,

$\displaystyle \sum_{k=1}^n (k+1)\left(k + \dfrac{1}{\omega}\right)\left(k + \dfrac{1}{\omega^2}\right)$

Notations: $\omega$ and $\omega^2$ are non-real cube roots of unity .

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\dfrac{n^2{(n+1)}^2}{4} - n

\dfrac{n^2{(n+1)}^2}{4} + n

\dfrac{n(n^2+2)}{3}

\dfrac{n(n^2-2)}{3}

2 solutions

Reynan Henry
Jan 4, 2017

notice that $\omega^3=1$ and $1+\omega+\omega^2=0$

$(k+1)(k+\frac{1}{\omega})(k+\frac{1}{\omega^2})=(k+1)(k+\omega^2)(k+\omega)=k^3+k^2(1+\omega+\omega^2)+k(\omega+\omega^2+\omega^3)+\omega^3=k^3+k^2\cdot 0 + k \cdot 0 \cdot \omega + 1 = k^3 +1$

so the problem can be written as $\sum_{k=1}^n k^3+1 = \frac{n^2(n+1)^2)}{4}+n$

Good solution indeed!

Tapas Mazumdar - 4 years, 5 months ago

Nice, same solutio

Jason Chrysoprase - 4 years, 4 months ago

Abhiram Bondada
Apr 2, 2017

Observe carefully, The given expression is the simplified form of equation with roots as negative cubeth roots of unity .

As 1/w=w^2 we can substitute this in the equation and thus get this result

Now this question will be sum of x^3 +1 from 1 to n and thats it use formula for sum of cubes of first n natural numbers and you are done

Complexity! (7)

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