若\(n\)為4的倍數，\(i=\sqrt{-1}\)，則級數\(S=1+2i+3i^2+\ldots+(n+1)i^n\)的和是多少？
ans：\(\displaystyle \frac{1}{2}(n+2-ni)\)
請大大們幫忙給出解題過程，謝謝！

Hint :
is = i + 2i^2 + 3i^3 + ..... + ni^n + (n+1)i^(n+1)
S - iS = ( 1 - i )S = .....