Question-113309

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

S_n =Σ_(k=0) ^(n ) n^(n−k) a^k ⇒S_n =n^n Σ_(k=0) ^n ((a/n))^k =n^n ×((1−((a/n))^(n+1) )/(1−(a/n))) =n^(n+1) ×((n^(n+1) −a^(n+1) )/(n^(n+1) (n−a))) =((n^(n+1) −a^(n+1) )/(n−a)) if a≠n if a=n S_n =Σ_(k=0) ^n n^k n^(n−k) =n^n Σ_(k=0) ^n (1) =(n+1)n^n .

$$\mathrm{S}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}\:} \:\mathrm{n}^{\mathrm{n}−\mathrm{k}} \:\mathrm{a}^{\mathrm{k}} \:\Rightarrow\mathrm{S}_{\mathrm{n}} =\mathrm{n}^{\mathrm{n}} \:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\left(\frac{\mathrm{a}}{\mathrm{n}}\right)^{\mathrm{k}} \:=\mathrm{n}^{\mathrm{n}} ×\frac{\mathrm{1}−\left(\frac{\mathrm{a}}{\mathrm{n}}\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{1}−\frac{\mathrm{a}}{\mathrm{n}}} \\ $$$$=\mathrm{n}^{\mathrm{n}+\mathrm{1}} ×\frac{\mathrm{n}^{\mathrm{n}+\mathrm{1}} −\mathrm{a}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}^{\mathrm{n}+\mathrm{1}} \left(\mathrm{n}−\mathrm{a}\right)}\:=\frac{\mathrm{n}^{\mathrm{n}+\mathrm{1}} −\mathrm{a}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}−\mathrm{a}}\:\:\:\mathrm{if}\:\mathrm{a}\neq\mathrm{n} \\ $$$$\mathrm{if}\:\mathrm{a}=\mathrm{n}\:\:\:\:\mathrm{S}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{n}^{\mathrm{k}} \:\mathrm{n}^{\mathrm{n}−\mathrm{k}} \:=\mathrm{n}^{\mathrm{n}} \:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \left(\mathrm{1}\right)\:=\left(\mathrm{n}+\mathrm{1}\right)\mathrm{n}^{\mathrm{n}} \:. \\ $$

Commented by Hassen_Timol last updated on 12/Sep/20

Thank you so much Sir

Commented by abdomsup last updated on 12/Sep/20

$${you}\:{are}\:{welcome} \\ $$

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

Σ_(k=1) ^n k.k!=1.1!+2.2!+.... 1.1!=(2−1)1!=2!−1! 2.2!=3.2!−2!=3!−2! .... Σ_(n=1) ^n k.k!=2!−1!+3!−2!+4!−3!+.....n!−(n−1)!+(n+1)!−n! =(n+1)!−1

$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}.{k}!=\mathrm{1}.\mathrm{1}!+\mathrm{2}.\mathrm{2}!+…. \\ $$$$\mathrm{1}.\mathrm{1}!=\left(\mathrm{2}−\mathrm{1}\right)\mathrm{1}!=\mathrm{2}!−\mathrm{1}! \\ $$$$\mathrm{2}.\mathrm{2}!=\mathrm{3}.\mathrm{2}!−\mathrm{2}!=\mathrm{3}!−\mathrm{2}! \\ $$$$…. \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{k}.{k}!=\mathrm{2}!−\mathrm{1}!+\mathrm{3}!−\mathrm{2}!+\mathrm{4}!−\mathrm{3}!+…..{n}!−\left({n}−\mathrm{1}\right)!+\left({n}+\mathrm{1}\right)!−{n}! \\ $$$$=\left({n}+\mathrm{1}\right)!−\mathrm{1} \\ $$

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

Σ_(k=0) ^n ((a^k /n^k ))n^n =n^n Σ_(k=0) ^n ((a/n))^k =n^n (((((a/n))^(n+1) −1)/((a/n)−1))) =n^(n+1) (((a^(n+1) −n^(n+1) )/(n^(n+1) (a−n))))=(((a^(n+1) −n^(n+1) )/(a−n)))

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

Commented by Hassen_Timol last updated on 12/Sep/20

Thank you so much

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