1+iw+(iw)^2 +(iw)^3 +.........(iw)^(989) =? ans= (2/(1−iw)) answer is correct. pls help .. how to do this? TIA


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 61915 by aseerimad last updated on 11/Jun/19

$1 + i w + {(i w)}^{2} + {(i w)}^{3} + . . . . . . . . . {(i w)}^{989} = ?$ $a n s = \frac{2}{1 - i w} a n s w e r i s c o r r e c t .$ $p l s h e l p . . h o w t o d o t h i s ?$ $T I A$
	Answered by tanmay last updated on 11/Jun/19

	$S = 1 + t + t^{2} + t^{3} + . . . + t^{n}$ $S = \frac{1 - t^{n + 1}}{1 - t} h e r e n = 989$ $n o w {(i w)}^{989 + 1} = {(i)}^{990} \times {(w)}^{990}$ $= {(i^{2})}^{495} \times {(w^{3})}^{330}$ $= {(- 1)}^{495} \times {(1)}^{330}$ $= - 1$ $s o a n s w e r i s = \frac{1 - (- 1)}{1 - i w} = \frac{2}{1 - i w}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum