i^(2 n + 1) = i for all positive integers n
Use M.I.:
Let n = 1:
i^(2n + 1) = i^3 = -i........
The question is not correct...... Do you mean i^(4n + 1) = i?
\(i^1=i\\ i^2=-1\\ i^3=-i\\ i^4=+1\\ i^5=i\\ ...\\ i^{1+4k}=i\)
2n+1=4k+1 where n and k are both positive integers
2n = 4k
n = 2k
n = 2,4,6, ..... all positive even integers