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Question Number 187437 by Shrinava last updated on 17/Feb/23

Answered by witcher3 last updated on 17/Feb/23

$$\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)\frac{1}{{\mathrm{m}}^{\mathrm{k}}}={(1+\frac{1}{\mathrm{m}})}^{\mathrm{n}}-1$$$$\iff 31+\underset{\mathrm{m}=1}{\sum ^{\mathrm{n}}}({\mathrm{m}}^{\mathrm{n}}({(1+\frac{1}{\mathrm{m}})}^{\mathrm{n}.}-1)={31}^{30}$$$$\iff 31+\underset{\mathrm{m}=1}{\sum ^{\mathrm{n}}}{(1+\mathrm{m})}^{\mathrm{n}}-{\mathrm{m}}^{\mathrm{n}}={31}^{30}$$$$\iff 31+{(\mathrm{n}+1)}^{\mathrm{n}}-1={31}^{30}\iff {(\mathrm{n}+1)}^{\mathrm{n}}={31}^{30}-30$$$$\mathrm{may}\mathrm{bee}$$$$1+\mathrm{\Sigma }(\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)\mathrm{\Sigma }{\mathrm{m}}^{\mathrm{n}-\mathrm{k}}={31}^{30}$$$$\iff {(\mathrm{n}+1)}^{\mathrm{n}}={31}^{30}\Rightarrow \mathrm{n}=30$$

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