Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 70132 by RAKESH MANDA last updated on 01/Oct/19

$$\underset{n=1}{\sum ^{3050}}{i}^n$$

Commented by mathmax by abdo last updated on 01/Oct/19

$$S=\sum _{n=0}^{3050}{i}^n-1=\frac{1-i^{3051}}{1-i}-1=\frac{1-i{\left(i^2\right)}^{1525}}{1-i}-1$$$$=\frac{1-i{(-1)}^{1525}}{1-i}-1=\frac{1+i}{1-i}-1=\frac{{(1+i)}^2}{2}-1$$$$=\frac{1+2i-1}{2}-1=i-1$$

Answered by naka3546 last updated on 01/Oct/19

$$i-1$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com