Q.

If α is a root of the equation x2+x+1=0 and ∑k=1n(αk+1αk)2=20, then n is equal to __________.          [2025]


Ans.

(11)

We have, α is root of equation x2+x+1=0.

Let α=ω.

Then, ∑k=1n(ωk+1ωk)2=∑k=1n(ω2k+1ω2k+2)

                                        =∑k=1n(ω2k+ωk+2)          (∵  ω3k=1)

But ∑k=1n(ω2k+ωk+2)=20

⇒ (ω2+ω4+...+ω2n)+(ω+ω2+...+ωn)+2n=20.

If n = 3m,  m ∈ I

Then, 0 + 0 + 2n = 20

⇒ n=10 (not satisfy as m is integer).

If n = 3m + 1, then ω2+ω+2n=20.

⇒ n=212          (not possible)

If n = 3m + 2, then (ω2+ω4)+(ω+ω2)+2n=20.

⇒ 2(ω2+ω+n)=20.

⇒ n=10+1=11.

∴   For n = 11 satisfies n = 3m + 2.