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Question Number 184037 by liuxinnan last updated on 02/Jan/23

$$i^2=-1$$$$\underset{j=1}{\sum ^{2023}}{ji}^j=?$$

Answered by SEKRET last updated on 02/Jan/23

$${\mathbf{i}}^2=-1$$$$\underset{\mathbf{j}=1}{\sum ^{2023}}\mathbf{j}\cdot {\mathbf{i}}^{\mathbf{j}}=$$$$\mathbf{i}=\mathbf{x}\mathbf{f}\left(\mathbf{x}\right)=1\cdot {\mathbf{x}}^1+2\cdot {\mathbf{x}}^2+3\cdot {\mathbf{x}}^3+.....+2023\cdot {\mathbf{x}}^{2023}$$$$\frac{\mathbf{f}\left(\mathbf{x}\right)}{\mathbf{x}}=1\cdot {\mathbf{x}}^0+2\cdot \mathbf{x}+3\cdot {\mathbf{x}}^2+...+2023\cdot {\mathbf{x}}^{2022}$$$$\mathbf{f}\left(0\right)=0\int \frac{\mathbf{f}\left(\mathbf{x}\right)}{\mathbf{x}}\mathbf{dx}=\mathbf{x}+{\mathbf{x}}^2+{\mathbf{x}}^3+...+{\mathbf{x}}^{2023}+\mathbf{c}$$$${\mathbf{b}}_1=\mathbf{x}\mathbf{q}=\mathbf{x}\mathbf{n}=2023{\mathbf{S}}_{\mathbf{n}}=\frac{\mathbf{x}({\mathbf{x}}^{2023}-1)}{\mathbf{x}-1}$$$$\int \frac{\mathbf{f}\left(\mathbf{x}\right)}{\mathbf{x}}\mathbf{dx}=\frac{\mathbf{x}\cdot ({\mathbf{x}}^{2023}-1)}{\mathbf{x}-1}+\mathbf{C}$$$$\frac{\mathbf{f}\left(\mathbf{x}\right)}{\mathbf{x}}=\frac{2023\cdot {\mathbf{x}}^{2024}-2024{\mathbf{x}}^{2023}+\mathbf{x}}{{\mathbf{x}}^2-2\mathbf{x}+1}$$$$\mathbf{f}\left(\mathbf{x}\right)=\frac{2023\cdot \mathbf{x}\cdot {\left({\mathbf{x}}^2\right)}^{1012}-2024\cdot {\left({\mathbf{x}}^2\right)}^{1012}+\mathbf{x}}{{\mathbf{x}}^2-2\mathbf{x}+1}$$$$\mathbf{f}\left(\mathbf{i}\right)=\frac{2023\mathbf{i}-2024+\mathbf{i}}{-2\mathbf{i}}=\frac{2024}{2}\cdot (-1-\mathrm{i})$$$$\mathbf{f}\left(\mathbf{x}\right)=1012(-1-\mathbf{i})$$$$\mathit{A}\mathit{B}\mathit{D}\mathit{U}\mathit{L}\mathit{A}\mathit{Z}\mathit{I}\mathit{Z}\mathit{A}\mathit{B}\mathit{D}\mathit{U}\mathit{V}\mathit{A}\mathit{L}\mathit{I}\mathit{Y}\mathit{E}\mathit{V}$$$$$$

Answered by liuxinnan last updated on 02/Jan/23

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