If {i^2} = - 1 , then sum i + {i^2} + {i^3} + ... to 1000 terms is equal to

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4-1.Complex numbers

easy

If ${i^2} = - 1$, then sum $i + {i^2} + {i^3} + ...$to $1000$ terms is equal to

$1$

$-1$

$i$

$0$

Solution

(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$.

Standard 11

Mathematics

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If ${i^2} = - 1$, then sum $i + {i^2} + {i^3} + ...$to $1000$ terms is equal to

Solution

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