Question

calculati 1+i+i^2+...+i^10

Answer 1

[tex]\text{Suma este o progresie geometrica de ratie i.}\\ 1+i+i^2+\ldots+i^{10}=\dfrac{i^{11}-1}{i-1}\\ \text{Sa observam ca:}\\ i^{4k}=1\\ i^{4k+1}=i\\ i^{4k+2}= -1\\ i^{4k+3}=-i,\forall k\in \mathbb{N}\\ 11:4=2,r=3\Rightarrow i^{11}=-i\\ \dfrac{i^{11}-1}{i-1}=\dfrac{-i-1}{i-1}=\dfrac{-(i+1)^2}{(i-1)(i+1)}=\dfrac{1-2i-1}{1+1}=\dfrac{-2i}{2}=\boxed{-i}[/tex]