Technical information
Behaviour being used: Adaptive mode
Minimum fraction: 0
Question summary: [P e n t r u space n equals # n comma space s a space s e space c a l c u l e z e space sum from k equals 1 to n of left parenthesis fraction numerator 1 minus i over denominator 1 plus i end fraction right parenthesis to the power of k] (paranteza din "suma" este la puterea k )
Right answer summary: The content can not be displayed.
Question state: todo