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Question Number 113309 by Hassen_Timol last updated on 12/Sep/20

Commented by Hassen_Timol last updated on 12/Sep/20

$$\mathrm{Can}\mathrm{you}\mathrm{please}\mathrm{help}\mathrm{me}...?$$

Commented by mathmax by abdo last updated on 12/Sep/20

$${\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{n}}^{\mathrm{n}-\mathrm{k}}{\mathrm{a}}^{\mathrm{k}}\Rightarrow {\mathrm{S}}_{\mathrm{n}}={\mathrm{n}}^{\mathrm{n}}\sum _{\mathrm{k}=0}^{\mathrm{n}}{\left(\frac{\mathrm{a}}{\mathrm{n}}\right)}^{\mathrm{k}}={\mathrm{n}}^{\mathrm{n}}\times \frac{1-{\left(\frac{\mathrm{a}}{\mathrm{n}}\right)}^{\mathrm{n}+1}}{1-\frac{\mathrm{a}}{\mathrm{n}}}$$$$={\mathrm{n}}^{\mathrm{n}+1}\times \frac{{\mathrm{n}}^{\mathrm{n}+1}-{\mathrm{a}}^{\mathrm{n}+1}}{{\mathrm{n}}^{\mathrm{n}+1}(\mathrm{n}-\mathrm{a})}=\frac{{\mathrm{n}}^{\mathrm{n}+1}-{\mathrm{a}}^{\mathrm{n}+1}}{\mathrm{n}-\mathrm{a}}\mathrm{if}\mathrm{a}\ne \mathrm{n}$$$$\mathrm{if}\mathrm{a}=\mathrm{n}{\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{n}}^{\mathrm{k}}{\mathrm{n}}^{\mathrm{n}-\mathrm{k}}={\mathrm{n}}^{\mathrm{n}}\sum _{\mathrm{k}=0}^{\mathrm{n}}\left(1\right)=(\mathrm{n}+1){\mathrm{n}}^{\mathrm{n}}.$$

Commented by Hassen_Timol last updated on 12/Sep/20

Thank you so much Sir

Commented by abdomsup last updated on 12/Sep/20

$$youarewelcome$$

Answered by Dwaipayan Shikari last updated on 12/Sep/20

$$\underset{k=1}{\sum ^n}k.k!=1.1!+2.2!+....$$$$1.1!=(2-1)1!=2!-1!$$$$2.2!=3.2!-2!=3!-2!$$$$....$$$$\underset{n=1}{\sum ^n}k.k!=2!-1!+3!-2!+4!-3!+.....n!-(n-1)!+(n+1)!-n!$$$$=(n+1)!-1$$

Answered by Dwaipayan Shikari last updated on 12/Sep/20

$$\underset{k=0}{\sum ^n}\left(\frac{a^k}{n^k}\right)n^n=n^n\underset{k=0}{\sum ^n}{\left(\frac{a}{n}\right)}^k=n^n\left(\frac{{\left(\frac{a}{n}\right)}^{n+1}-1}{\frac{a}{n}-1}\right)$$$$=n^{n+1}\left(\frac{a^{n+1}-n^{n+1}}{n^{n+1}(a-n)}\right)=\left(\frac{a^{n+1}-n^{n+1}}{a-n}\right)$$

Commented by Hassen_Timol last updated on 12/Sep/20

Thank you so much

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