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Question Number 160753 by Gbenga last updated on 05/Dec/21

Commented by Tinku Tara last updated on 06/Dec/21

$$\mathrm{Please}\mathrm{rotate}\mathrm{image}\mathrm{before}\mathrm{uploading}$$

Answered by qaz last updated on 06/Dec/21

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }(\underset{\mathrm{j}=1}{\sum ^{\mathrm{k}}}\frac{{(\mathrm{n}+\mathrm{j})}^{\mathrm{m}}}{{\mathrm{n}}^{\mathrm{m}-1}}-\mathrm{kn})$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }(\underset{\mathrm{j}=1}{\sum ^{\mathrm{k}}}\mathrm{n}{(1+\frac{\mathrm{j}}{\mathrm{n}})}^{\mathrm{m}}-\mathrm{kn})$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }(\underset{\mathrm{j}=1}{\sum ^{\mathrm{k}}}\mathrm{n}(1+\frac{\mathrm{mj}}{\mathrm{n}}+\mathrm{o}\left(\frac{1}{\mathrm{n}}\right))-\mathrm{kn})$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }(\mathrm{kn}+\frac{(\mathrm{k}+1)\mathrm{km}}{2}-\mathrm{kn}+\mathrm{o}\left(\frac{1}{\mathrm{n}}\right))$$$$=\frac{(\mathrm{k}+1)\mathrm{km}}{2}$$

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