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Question Number 185580 by Shrinava last updated on 23/Jan/23

Answered by JDamian last updated on 24/Jan/23

$$S=\underset{k=1}{\sum ^n}[\left(\begin{array}{c}n\\ k\end{array}\right)\cdot \underset{m=1}{\sum ^n}{m}^{n-k}]$$$$=\underset{k=1}{\sum ^n}\underset{m=1}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)\cdot m^{n-k}=\underset{m=1}{\sum ^n}\underset{k=1}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)\cdot m^{n-k}$$$$=\underset{m=1}{\sum ^n}[{(m+1)}^n-m^n]={(n+1)}^n-1$$$$$$$$31+{(n+1)}^n-1={31}^{30}$$

Commented by Shrinava last updated on 24/Jan/23

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor},\mathrm{to}\mathrm{be}\mathrm{continued}?$$

Commented by JDamian last updated on 25/Jan/23

$$31+{(n+1)}^n-1={31}^{30}$$$${(n+1)}^n-1={31}^{30}-31=31({31}^{29}-1)$$$$[{(n+1)}^n-1]\mathrm{mod}31=31({31}^{29}-1)\mathrm{mod}31=0$$$$[{(n+1)}^n-1]\mathrm{mod}31=0$$

Commented by Shrinava last updated on 25/Jan/23

$$\mathrm{Ans}=0\mathrm{dear}\mathrm{professor}?$$

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