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STD 11 - 4.1 complex nubers — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 4.1 complex nubers1 Mark
MCQ
If ${i^2} = - 1$, then sum $i + {i^2} + {i^3} + ...$to $1000$ terms is equal to
  • A
    $1$
  • B
    $-1$
  • C
    $i$
  • ✓
    $0$
✓

Answer

Correct option: D.
$0$
d
(d) $i + {i^2} + {i^3} + .......$up to $1000$ terms
$ = \frac{{i(1 - {i^{1000}})}}{{1 - i}} = $$\frac{{i(1 - {{({i^4})}^{250}})}}{{1 - i}}$$ = \frac{{i(1 - 1)}}{{1 - i}} = 0$.

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