If$\alpha is a root of the equation$x + x + 1 = 0$and$$\sum 2 n ($$\alpha$k + 1$)$2 = 20$, then n is equal to$k=1$k$$\alpha$
Step-by-Step Solution
Key Concept: Use$\alpha^3=1$and$1+\alpha+\alpha^2=0$to reduce the summation cyclically.
$\alpha =$$\omega 2 (11)$∴ ($\omega$k + 1$) =$$\omega$2k + 1 + 2$k 2k$$\omega$$\omega 2k k 3k =$$\omega +$$\omega + 2$∵$\$omega = 1$n 2k k$∴$\sum ($$\omega +$$\$omega + 2) = 20$$k=1$2 4 6 2n 2 3$$\Rightarrow$ ($\omega +$$$\omega +$$$\omega +$$$\ldots +$$$\omega ) + ($$$\omega +$$$\omega +$$$\omega +$$$\ldots+ n$$$\omega ) +$$2n = 20$Now if$n = 3m$, m$$\i_n I Then$0 + 0 + 2n = 2$0$$\Rightarrow$$n = 10$(not satisfy) if$n = 3m + 1$, then 2$\omega +$$\$omega + 2n = 20$$21 - 1 + 2n = 2$0$\Rightarrow $n = ($not possible ) 2 if$n = 3m + 2$, 8 10 4 5 ($$\omega +$$$\omega ) + ($$$\omega +$$$\omega ) +$2n = 20$2 2$\Rightarrow ($\omega +$$$\omega) + ($$$\omega +$$$\omega ) +$$2n = 20$$2n = 22$$n = 11$satisfy$n = 3$m + 2$∴$n = 11$
Correct Answer: 11