Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

Answered by JDamian last updated on 24/Jan/23

S=Σ_(k=1) ^n [ ((n),(k) ) ∙Σ_(m=1) ^n m^(n−k) ]     =Σ_(k=1) ^n Σ_(m=1) ^n  ((n),(k) )∙m^(n−k) =Σ_(m=1) ^n Σ_(k=1) ^n  ((n),(k) )∙m^(n−k)      =Σ_(m=1) ^n [(m+1)^n −m^n ]=(n+1)^n −1    31+(n+1)^n −1=31^(30)

$${S}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:\centerdot\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}{m}^{{n}−{k}} \right] \\ $$$$\:\:\:=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\centerdot{m}^{{n}−{k}} =\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\centerdot{m}^{{n}−{k}} \\ $$$$\:\:\:=\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\left({m}+\mathrm{1}\right)^{{n}} −{m}^{{n}} \right]=\left({n}+\mathrm{1}\right)^{{n}} −\mathrm{1} \\ $$$$ \\ $$$$\mathrm{31}+\left({n}+\mathrm{1}\right)^{{n}} −\mathrm{1}=\mathrm{31}^{\mathrm{30}} \\ $$

Commented by Shrinava last updated on 24/Jan/23

thank you dear professor, to be continued?

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

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]mod 31=31(31^(29) −1)mod 31=0  [(n+1)^n −1]mod 31=0

$$\mathrm{31}+\left({n}+\mathrm{1}\right)^{{n}} −\mathrm{1}=\mathrm{31}^{\mathrm{30}} \\ $$$$\left({n}+\mathrm{1}\right)^{{n}} −\mathrm{1}=\mathrm{31}^{\mathrm{30}} −\mathrm{31}=\mathrm{31}\left(\mathrm{31}^{\mathrm{29}} −\mathrm{1}\right) \\ $$$$\left[\left({n}+\mathrm{1}\right)^{{n}} −\mathrm{1}\right]\mathrm{mod}\:\mathrm{31}=\mathrm{31}\left(\mathrm{31}^{\mathrm{29}} −\mathrm{1}\right)\mathrm{mod}\:\mathrm{31}=\mathrm{0} \\ $$$$\left[\left({n}+\mathrm{1}\right)^{{n}} −\mathrm{1}\right]\mathrm{mod}\:\mathrm{31}=\mathrm{0} \\ $$

Commented by Shrinava last updated on 25/Jan/23

Ans = 0 dear professor?

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com