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The value of `sum_(n=0)^(100)i^(n!)` equals (where `i=sqrt(-1))`A. `-1`B. `i`C. `2i+95`D. `97+i` |
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Answer» Correct Answer - C `(c )` `S=sum_(n=0)^(100)(i)^(n!)` `=(i)^(0!)+(i)^(1!)+(i)^(2!)+…..` `=i+i-1+i^(6)+i^(24)+(i)^(5!)+(i)^(6!)+..+(i)^(100!)` `=95+2i` |
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