[JEE Advanced 1996] 1 2 omega 2 omega 2 2 3 omega 3 omega 2 n 1
$1\cdot(2-\omega)(2-\omega^2)+2\cdot(3-\omega)(3-\omega^2)+\cdots+(n-1)(n-\omega)(n-\omega^2)$ ($\omega$ a cube root of unity) equals ____
1 Answer
VAVidaara Admin
✓ Vidaara Team
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· 2d ago
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Answer: $\dfrac{n^2(n+1)^2}4-n$.
$(k-\omega)(k-\omega^2)=k^2+k+1$, so the $k$-th term is $(k-1)(k^2+k+1)=k^3-1$. Summing $k=2$ to $n$: $\big(\frac{n(n+1)}2\big)^2-1-(n-1)=\frac{n^2(n+1)^2}4-n$.
JEE Advanced 1996 · Complex Numbers — verified solution by the Vidaara Team.
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