Complex Analysis 2

Algebra Level 4

( 1 + 1 ω ) ( 1 + 1 ω 2 ) + ( 2 + 1 ω ) ( 2 + 1 ω 2 ) + … + ( 99 + 1 ω ) ( 99 + 1 ω 2 ) + ( 100 + 1 ω ) ( 100 + 1 ω 2 ) \large\left(1 + \frac1\omega\right)\left(1 + \frac1{\omega^2}\right) + \left(2 + \frac1\omega\right)\left(2 + \frac1{\omega^2}\right) + \ldots \\ \large + \left(99+ \frac1\omega\right)\left(99 + \frac1{\omega^2}\right)+ \left(100 + \frac1\omega\right)\left(100 + \frac1{\omega^2}\right)

Let ω \omega denote the complex cube root unity. Evaluate the summation above.


The answer is 333400.

Surya Prakash Bugatha by Surya Prakash Bugatha , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Daniel Liu
Jun 27, 2015

Note that because ω 3 = 1 \omega^3=1 and ω 2 + ω = − 1 \omega^2+\omega = -1 , we have ( k + 1 ω ) ( k + 1 ω 2 ) = ( k + ω 2 ) ( k + ω ) = k 2 + ( ω 2 + ω ) k + 1 = k 2 − k + 1 \begin{aligned}\left(k+\dfrac{1}{\omega}\right)\left(k+\dfrac{1}{\omega^2}\right)&=\left(k+\omega^2\right)\left(k+\omega\right)\\ &= k^2+(\omega^2+\omega)k+1\\ &= k^2-k+1 \end{aligned}

Thus our answer is ∑ k = 1 100 k 2 − k + 1 = 100 ⋅ 101 ⋅ 201 6 − 100 ⋅ 101 2 + 100 = 333400 \sum_{k=1}^{100}k^2-k+1=\dfrac{100\cdot 101\cdot 201}{6}-\dfrac{100\cdot 101}{2}+100=\boxed{333400}

11 Helpful 0 Interesting 0 Brilliant 0 Confused

I did same. (+1)

Dev Sharma - 5 years, 5 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...