For any integer k, let `alpha_(k) = cos(kpi)/(7) + isin.(kpi)/(7)`, where `I = sqrt(-1)`. Value of the expression.`(sum_(k=1)^(12)|alpha_(k+1)-alpha_(k)|)/(sum_(k=1)^(3) |alpha_(4k-1)-alpha_(4k-2)|) "is"`________.

Answer

4

Explanation

`alpha_(k) =cos.(kpi)/(7) +isin .(kpi)/(7) = e^((kpi)/(7)i)`
`therefore (underset(k=1)overset(12)sum|alpha_(K+1)-alpha_(k)|)/(sum_(k-=1)^(3)|alpha_(4k-1)-alpha_(4k-2)|)` ` = (sum_(k=1)^(12)|e^((k+1)/(7))-e^((kpi)/(7)i)|)/(sum_(k=1)^(3) |e^(4k-1)/(7)i-e^((4k-2pi)/(7)i)|)`
`=(sum_(k=1)^(12) |e^((kpi)/(7)i)||e^((pii)/(7))-1|)/(sum_(k=1)^(3)|e^(((4k -2)pi)/(7)i)||e^(pii)/(7)-1|)`=`(sum_(k=1)^(12))/(sum_(k=1)^(2)) " "(because |e^((kpi)/(7)i)| =|e^(((4k-2)pi)/(7)i)|=1)`
`=(12)/(3) =4`