Σlimits_{n - 1}^{50} {{i^{(2n - 1)!}}} is equal to (where i = √ {-1} )

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4-1.Complex numbers

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$\sum\limits_{n - 1}^{50} {{i^{(2n - 1)!}}} $ is equal to (where $i = \sqrt {-1}$ )

A

$48$

B

$48 + i$

C

$47 + i$

D

$48+2i$

Solution

${i^{1!}} + {i^{3!}} + {i^{5!}} + {i^{7!}} + …. + {i^{99!}}$

$ = i + {i^6} + 1 + 1 + ……..1$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \leftarrow 48\,times$

$ = i – 1 + 48 = 47 + i$

Standard 11

Mathematics

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