Question

Mark the correct alternative in the folowing question:

$$\begin{gathered} {\left(-1\right)}^{301}+{\left(-1\right)}^{302}+{\left(-1\right)}^{303}+...+{\left(-1\right)}^{400} \\ \left(\mathrm{a}\right)1\left(\mathrm{b}\right)101\left(\mathrm{c}\right)100\left(\mathrm{d}\right)0 \end{gathered}$$

Answer 1

Since,

$$\begin{gathered} {\left(-1\right)}^{301}+{\left(-1\right)}^{302}+{\left(-1\right)}^{303}+...+{\left(-1\right)}^{400} \\ =\left(-1\right)+\left(1\right)+\left(-1\right)+...+\left(1\right)\left[\mathrm{As},{\left(-1\right)}^{\mathrm{odd}}=-1\mathrm{and}{\left(-1\right)}^{\mathrm{even}}=1\right] \\ =-50+50\left[\mathrm{As},\mathrm{there}\mathrm{are}50\left(-1\right)\text{'}\mathrm{s}\mathrm{and}50\left(1\right)\text{'}\mathrm{s}\right] \\ =0 \end{gathered}$$

Hence, the correct alternative is option is (d).