MCQ
$(-1)^{301}+(-1)^{302}+(-1)^{303}+\dots+(-1)^{400}$
  • A
    $1$
  • B
    $101$
  • C
    $100$
  • $0$

Answer

Correct option: D.
$0$

Since,
$(-1)^{301}+(-1)^{302}+(-1)^{303}+\dots+(-1)^{400}$
$=(-1)+(1)+(-1)+\dots+(1)$ $\Big[\text{As, }(-1)^{\text{odd}}=-1\text{ and }(-1)^{\text{even}}=1\Big]$
$=-50+50$ $[\text{As, there are} 50(-1)'\text{s and 50 (1)}'\text{s}]$
$=0$
Hence, the correct alternative is option is $(d)$.

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