summation k=1 to 2015 of
1/i^(4m+k), m belongs to N

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have}, \\ \mathrm{S}=\sum _{\mathrm{k}=1}^{1500} \frac{1}{{\mathrm{i}}^{4\mathrm{m}+\mathrm{k}}} \\ =\sum _{\mathrm{k}=1}^{1500} \frac{1}{{\mathrm{i}}^{4\mathrm{m}}.{\mathrm{i}}^{\mathrm{k}}} \\ \mathrm{We} \mathrm{know} \mathrm{that} {\mathrm{i}}^{4\mathrm{n}}=1 \mathrm{if} \text{'}\mathrm{n}\text{'} \mathrm{is} \mathrm{integer}. \\ \Rightarrow {\mathrm{i}}^{4\mathrm{m}}=1 \\ \Rightarrow \mathrm{S}=\sum _{\mathrm{k}=1}^{1500} \frac{1}{{\mathrm{i}}^{\mathrm{k}}} \\ =\sum _{\mathrm{k}=1}^{1500} {\left(\frac{1}{\mathrm{i}}\right)}^{\mathrm{k}} \\ =\sum _{\mathrm{k}=1}^{1500} {\left(\frac{\mathrm{i}}{{\mathrm{i}}^2}\right)}^{\mathrm{k}} \\ =\sum _{\mathrm{k}=1}^{1500} {\left(-\mathrm{i}\right)}^{\mathrm{k}} \left[\mathrm{As} {\mathrm{i}}^2=-1\right] \\ =\left(-\mathrm{i}\right)+{\left(-\mathrm{i}\right)}^2+{\left(-\mathrm{i}\right)}^3+.....{\left(-\mathrm{i}\right)}^{1500} \\ \mathrm{Which} \mathrm{is} \mathrm{G}.\mathrm{P}. \mathrm{of} 1500 \mathrm{terms} \mathrm{having} \mathrm{first} \mathrm{term}\left(\mathrm{a}\right)=-\mathrm{i} \mathrm{and} \mathrm{common} \mathrm{ratio}\left(\mathrm{r}\right)=-\mathrm{i} \\ \Rightarrow \mathrm{S}=\mathrm{a}\left(\frac{{\mathrm{r}}^{1500}-1}{\mathrm{r}-1}\right) \\ =\left(-\mathrm{i}\right)\left\{\frac{{\left(-\mathrm{i}\right)}^{1500}-1}{-\mathrm{i}-1}\right\} \\ =\left(-\mathrm{i}\right)\left\{\frac{{\left(\mathrm{i}\right)}^{1500}-1}{-\mathrm{i}-1}\right\} \\ \mathrm{As} 1500 \mathrm{is} \mathrm{a} \mathrm{multiple} \mathrm{of} 4 \mathrm{and} {\mathrm{i}}^{4\mathrm{n}}=1, \mathrm{hence} \\ \mathrm{S}=\left(-\mathrm{i}\right)\left\{\frac{1-1}{-\mathrm{i}-1}\right\}=\left(-\mathrm{i}\right)\left\{\frac{0}{-\mathrm{i}-1}\right\} \\ \Rightarrow \mathrm{S}=0 \\ \mathbf{\Rightarrow }\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{1}}^{\mathbf{1500}}\mathbf{ }\frac{\mathbf{1}}{{\mathbf{i}}^{\mathbf{4}\mathbf{m}\mathbf{+}\mathbf{k}}}\mathbf{=}\mathbf{0}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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